第二章实数(讲义)(解析版)_2026语数英7-9年级上《举一反三系列》多版本_2026《初中数学•举一反三系列》8上多版本_2026《初中数学•举一反三系列》8上(北师大)_第二章实数(讲义)

第二章  实数（举一反三讲义）全章题型归纳
 2024
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 1   ! ........................................................................................................................5
 2  "# !$% ............................................................................................................7
 3  & !' ........................................................................................................................9
 4  ()*+, ..........................................................................................................................................11
 5  -.)*/, ......................................................................................................................................14
 6  012*3456789*:; ..................................................................................................15
 7  "#012*<=>? ..................................................................................................................17
 8  @?012 ......................................................................................................................................18
 9  012A*B)' ......................................................................................................................20
 10 012*+,C>?' ..............................................................................................................22
DE .................................................................................................................................................................24
 11 012*FGHI< ......................................................................................................................24
 12 ()*JKL ..................................................................................................................................26
 13 M)NOPJ)NO ......................................................................................................................28
 14 ()Q)RST+# ..........................................................................................................................30
 15 ()+,*U# ..................................................................................................................................33
 16 VT012*>? ..........................................................................................................................36
 17 "#OW7.>' ..........................................................................................................................38
 18 012*XYZ[ ..........................................................................................................................44
 19 012*\]9 ..............................................................................................................................47
 20 012*(^U# ..........................................................................................................................50
知识点1  算术平方根和平方根的区别与联系


	



 x 
 a, x²=a
 x
 a 



 χ 
 a, x²=a
 x 
 a  
!
"#


$%


%&
'()
*+,-./%
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012
 a * 𝑎
 a 01* ±
𝑎
3456
𝑎7%89:.;

𝑎≥0
𝑎≥0
a <<.
<* 0
=>
?@A
BC
D
=>
E 0
0 FGB 0
知识点2  开平方
1.	HI
 a J
K
a LK.
2.KF"#M=>
1)KN
LK a OPB:..
2)B
BKQ-KB
RJ
BIST.
3)FK)*UJ
<VWJXYZKQB[\.
知识点3  
𝐚²与𝐚𝟐的性质
]^
;_
1`
𝒂²
𝒂𝟐=|𝒂|= 𝒂_𝒂≥𝟎`
−𝒂_𝒂< 𝟎`
62=|6|=6
(−6)2=|−6|=6
𝒂𝟐
( 𝐚)𝟐 = 𝐚_𝐚≥𝟎`
( 6)
2 =6
知识点4  立方根和平方根的不同点和相同点
a

	



 x a
 a,𝑥3=aa x
 a ab
 !



 x 
 a, x²=a, x 
 a  
!

c
 a %d$%
a

aB
.
aB.

%&
'(
)*+,-./%
012
3 𝑎
±
𝑎
"#
𝑎3456
ef
𝑎≥0
+gh
E 0
0 B 0
0 aB 0
知识点5  开立方
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1.bc) a *!*+,def!a6A a degf).
2.fPf!*hi
f
f!
+,j	
±
3
gf)
:.
k8)
c)
0 $%
-

%&-./%
efaG$%


知识点6  无理数
-lmnoJ)de-.)p
-.)*qrs27tuvw
_1`ffmx*)*yUz-.)a{ 2a
3 5|}
_2`~€π5b‚7π*),{ 2πa
π
2|}
_3`t-lmnoJ)s2ƒ„*),{ 0.101 001 000 1…_…c 1 †‡ˆ1‰bc 0)|p8-.)*J
)NOŠ)-l}-.)*J)NOmno}-.)m‹ŒŽO)*s2p
知识点7  实数的概念及分类
1. 34
7.)P-.)‘()p
2. O’
()7…wO’“”
_1`•]9O’
()–O‘7.)P-.)p
_2`•—I<O’
()–O‘—() 0 I()p
—M)
—O)
IM)
IO)
()
—()
I()
—7.)
—-.)
I7.)
0
()
7.)
0
-.)
—7.)
I7.)
—-.)
I-.)
7lJ)CnoJ)
-lmnoJ)
—M)
—O)
IM)
IO)
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知识点8  实数与数轴的关系
™bc()š–t#)R›*bcœŒ}žŸa)R›*™bcœšŒbc()p()Q)R›
*œbb U.
知识点9  实数范围内的有关概念
¡
<=
¢£
yž)
¤ a Q b ¥‘yž)a¦𝑎+ 𝑏= 0
3*yž)z−3
§)
¤ a Q b ¥‘§)a¦𝑎𝑏= 1
2 *§)z
1
2
k¨()*© 'šzHI)aª|𝑎| ≥0
|𝑎|
𝑎(𝑎> 0)
0(𝑎= 0)
−𝑎(𝑎< 0)
© '
¥‘yž)*…c)*© 'y|aª|𝑎| = |−𝑎|
| 2| = |−2| =
2
知识点10  实数的运算
«()¬­®am¯–t°±² ³ ´ µ_µ)m‘¶` ´+,a·¸–t°±f!+,t5
HI()*f+,p
7.)*+,<=P+,Y«()¬­®¹º»#a()¼T+,*½¾Q7.)¼T+,*½¾y¿a
À´ faÁ´µa@²³p¿Ã+,•ÄÅÆÇ*½¾°±a7È	*À,È	ÉÊ*p
知识点11  实数的大小比较
7.)JKL*Ë«()¬­®¹º»#p
…()*JÌ{u
—()šÍ 0aI()šJÍ 0a—()ÍbÎI()}…c—()a© '
*—()}…cI()a© '*I()J}«)R›Œ*…c()aÇÏ*)ÐKÅÏ*)
pÑÒaKL…()*JÓ7{uË
_1`ԟKL…()**Ja°·Õ]()*JÌp{KL 10Q 3 *Ja&Í( 10)
2 = 10
，32 = 9，10 > 9aÖ 10 > 3p
_2`#/,*Ë„-.)*×Ø'ÂaÁKL…)*Jp
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_3`مcÚ	*-.)KLJÛa–U#{uÜÝ
① 𝑎> 𝑏≥0⇔𝑎>
𝑏p
② 𝑎> 𝑏⇔
3 𝑎>
3 𝑏p
【培优篇】
【题型1  直接求平方根、立方根】
£ 1_24-25 ÞßÃu·àáâã·äå`uæçËAèé*z_    `
Ap 5z 5 *,ê
Bp0 *P!šz 0
Cp 9*z±3
Dp−3z(−3)2*bc
ëìC
OíîïðQ,ê*34añòbOíóôõ*—Õ<p
ö$$
ôõ A
 5z 5 *,êa
,êzHI)a¸ 5÷ø( 5)
2
= 5aÖ A —Õ}
ôõ B
0 *P!šz 0a
P!*]9Aa0 *P!ù‘ 0aÖ B —Õ}
ôõ C
 9*z ± 3a
9 = 3a· 3 *U‘ ±
3a·H ± 3a ± 3*z 9aúÍ 9*¼ûèéaÖ C èé}
ôõ D
−3z(−3)2*bca
(−3)2 = 9a9 *‘ ± 3aüÑ−3z 9 *bcaÖ D —Õ}
Öô Cp
ý2 1-1_24-25 ÞßÃu·þÿĀáā·Ă㥹`−
8
27*!z   } 4*,êz     p
ëì     
−
2
3     
2
Oíîïð!a,êa$*ĆzćĈ
{ĉbc—)𝑥*|Í𝑎aĊċČc—)𝑥
de𝑎*,ê}{ĉbc)𝑦*!|Í𝑏aĊċČc)𝑦de𝑏*!pčÑ$몖p
ö$$
−
8
27*!z−
2
3}
∵4 = 2a2*,êz 2a
∴4*,êz 2p
Öëì‘
−
2
3} 2p
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ý2 1-2_24-25 ÞßÃu·Ďď·ĐäđĒ`_1`ēŒ
a
0.000001
0.001
1
1000
1000000
3 𝑎
_2`čĔĕĖ*XYēė
①
3 3 = 1.442a¦
3 3000 = ______a
3 0.003 = ______}
②Ęę
3 343 = 7a¦
3 0.000343 = ______p
ëì_1`0.01a0.1a1 a10 a100_2`①14.42a0.1442a ②0.07
OíîĚěïðĜ!*<=aˆčgf)J)œĝÅCĝÇĞğ 3 Š U*!*J)œ
ĝÅCĝÇĞğ 1 Š$ª–aĠĄćĈgf)J)œQ U*!J)œĞğXYz$*Ćp
_1`"#!*<=$ª–}
_2`①"#!*<=$ª–}
②"#!*<=$ª–p
ö$$
_1`
3 0.000001 = 0.01}
3 0.001 = 0.1}
3 1 = 1}
3 1000 = 10}
3 1000000 = 100}
Öëì‘
0.01a0.1a1 a10 a100}
_2`①
3 3000 =
3 1000 ×
3 3 = 10 × 1.442 = 14.42}
3 0.003 =
3 0.001 ×
3 3 = 0.1 × 1.442 = 0.1442}
Öëì‘
14.42a0.1442}
②
3 0.000343 =
3 0.000001 ×
3 343 = 0.01 × 7 = 0.07
Öëì‘
0.07p
ý2 1-3¤ a b ¥‘yž)ac d ¥‘I§)a¦ 𝑎2−𝑏2 +
3 𝑐𝑑=       
.
ëì―1
ö$č8ġ
a+b=0acd=-1a
¦ 𝑎2−𝑏2 +
3 𝑐𝑑= (𝑎+ 𝑏)(𝑎−𝑏) +
3 −1 =
0−1=-1.
Öëìz:-1.
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【题型2  利用求平方根、立方根解方程】
£ 2_25-26 ĢßÛ·Ďď·ģÂđĒ` Í()𝑎,𝑏a]9bw+,“ ⊕”
𝑎⊕𝑏= 𝑎2−2𝑏pÍ𝑥*
%(𝑥+ 1) ⊕ 8 = 0*$‘            
p
ëì𝑥1 = 3,𝑥2 = −5
Oíč8]9*\+,aæ„Í𝑥*bĤ01%ª–a$†ª–ġÆëì.
ö$&8–ġ(𝑥+ 1)2−16 = 0a
∴(𝑥+ 1)2 = 16,
∴𝑥+ 1 = 4C𝑥+ 1 = −4,
∴𝑥1 = 3,𝑥2 = −5p
Öëì‘
𝑥1 = 3,𝑥2 = −5.
œĥîïð*zbĤ01%$*Ħa()*+,P.$]9*\+,z$*Ć.
ý2 2-1_24-25 ÞßÃu·ħāĨĩ·Ă㥹`uæó2A* xp
(1)𝑥2 = 81}
(2)(𝑥+ 1)3 = −27p
ëì(1)𝑥=± 9}
(2)𝑥= −4}
Oí_1`č*]9$𝑥*'abc—)7…caĪī¥‘yž)p
_2`č!*]9$𝑥*'aÀ„−27*!aÁġ„𝑥+ 1*'a°·„𝑥p
îĚěïðĜP!*]95U#pĠĄćĈP!_{ĉbc)*!|Í𝑎aĊċ
Čc)de𝑎*!`*34z$*Ćp
ö$_1`$
𝑥2 = 81a 
𝑥=±
81
𝑥=± 9}
_2`$
(𝑥+ 1)3 = −27a 
𝑥+ 1 =
3 −27a
𝑥+ 1 = −3a
∴x = −4p
ý2 2-2_24-25 ÞßÃu·ĬāĭĮį·äå`÷ø%(2𝑥−1)2−16 = 9* x *'‘      
p
ëì3 C−2
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OíîïðĜ"#$%p
ÀM.ġÆ(2𝑥−1)2 = 25a¦2𝑥−1 = 5C2𝑥−1 = −5aİ,ª–p
ö$$
∵(2𝑥−1)2−16 = 9
∴(2𝑥−1)2 = 25
∴2𝑥−1 = 5C2𝑥−1 = −5
∴𝑥= 3C𝑥= −2p
Öëì‘
−2C 3p
ý2 2-3_24-25 ÞßÃu·ıIJ·äA`uæ%A x *'
(1)2𝑥2 = 50}
(2)
1
3(𝑥+ 1)2 = 3}
(3)(2−𝑥)3 = 64}
(4)64(𝑥+ 1)3−125 = 0
ëì(1)𝑥= 5C𝑥= −5
(2)𝑥= 2C𝑥= −4
(3)𝑥= −2
(4)𝑥=
1
4
OíîĚěïðĜ"#P!iję)*'aĠĄćĈ$bc)*5!z$
*Ćp
_1`č*]9$ª–}
_2`č*]9$ª–}
_3`"#!*]9$ª–}
_4`"#!*]9$ª–p
ö$_1`$
2𝑥2 = 50
𝑥2 = 25a
$ġ
𝑥= 5C𝑥= −5}
_2`$
1
3(𝑥+ 1)2 = 3
(𝑥+ 1)2 = 9a
𝑥+ 1 = 3C𝑥+ 1 = −3
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$ġ
𝑥= 2C𝑥= −4}
_3`$
(2−𝑥)3 = 64
2−𝑥= 4a
$ġ
𝑥= −2}
_4`$
64(𝑥+ 1)3−125 = 0
(𝑥+ 1)3 =
125
64 a
𝑥+ 1 =
5
4a
$ġ
x = 1
4p
【题型3  由平方根、立方根求值】
£ 3_24-25 ÞßÃu·àáĴĵ·Ă㥹`Ęę|𝑎−6|Q 𝑎+ 2𝑏¥‘yž)a𝑐+ 5*!z 2a¦
𝑎−2𝑏−𝑐*‘     p
ëì ± 3
OíÑĚěïðĜyž)*]9a!aa—ÕćĈy]9z$ĆpÀ"#! 
¥‘yž)*]9ġ„𝑎a𝑏a𝑐*'}Ķķ$ġ„𝑎−2𝑏−𝑐*'aÁ$ª–p
ö$$
∵|𝑎−6|Q 𝑎+ 2𝑏¥‘yž)a
∴|𝑎−6| +
𝑎+ 2𝑏= 0a
∴𝑎−6 = 0a𝑎+ 2𝑏= 0a
$ġ
𝑎= 6a𝑏= −3a
∵𝑐+ 5*!z 2a
∴𝑐+ 5 = 8a
∴𝑐= 3
∴𝑎−2𝑏−𝑐= 6−2 × (−3)−3 = 9
∴𝑎−2𝑏−𝑐*z ± 3p
Öëì‘
 ± 3p
ý2 3-1Ęę x *…cz𝑎+ 3Q2𝑎−15a¸2𝑏−1*,êz 3p
(1)𝑎,𝑏,𝑥*'}
(2)𝑎+ 𝑏−1*!p
ëì(1)4a5a49
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(2)2
Oíîïða,ê5!*]9pĠĄćĈ6]95<=z$*Ćp
_1`čQ,ê*]9ª–ġ𝑎a𝑏*'aÁ$𝑥*'ª–}
_2`ĸ𝑎a𝑏*'Ķķ𝑎+ 𝑏−1Aİ,Â"#!*]9ª–ġëìp
ö$_1`$
 ∵𝑥*…cz𝑎+ 3Q2𝑎−15a¸2𝑏−1*,êz 3a
∴𝑎+ 3 + 2𝑎−15 = 0a2𝑏−1 = 9a
$ġ
𝑎= 4a𝑏= 5}
∴𝑥= (𝑎+ 3)2 = 72 = 49}
_2`$
 ∵𝑎= 4a𝑏= 5a
∴𝑎+ 𝑏−1 = 4 + 5−1 = 8
ý2 3-2_24-25 ÞßÃu·ĹĺĻÅ·Ă㥹`Ęę3𝑎+ 1*…cOiz7−𝑚P2−2𝑚a9 + 𝑏*
!z 2p
(1)𝑚a𝑎a𝑏*'}
(2) 7𝑎−𝑏*p
ëì(1)𝑚= 3a𝑎= 5a𝑏= −1
(2) ±
6
OíÑïða!aĠĄćĈyęļz$*Ć}
_1`č]9a!]9æġ7−𝑚+ 2−2𝑚= 0a9 + 𝑏= 23aª–„𝑚a𝑎a𝑏*'}
_2`À„7𝑎−𝑏*'aÁ"#]9„ë쪖p
ö$_1`$
ˆ8ġa7−𝑚+ 2−2𝑚= 0 
$ġ𝑚= 3        
Ö7−𝑚= 4,2−2𝑚= −4
∴3𝑎+ 1 = 42a
$ġ𝑎= 5 
&89 + 𝑏= 23a
$ġ𝑏= −1}
_2`∵7𝑎−𝑏= 7 × 5−(−1) = 36a
36 = 6 a6 *‘ ±
6 a
Ľt 7𝑎−𝑏*‘ ±
6p
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ý2 3-3_24-25 ÞßÃu·ľĺ\Ŀ·äå`Ęę2𝑎+ 1*z ± 3.3𝑎+ 2𝑏−4*!z−2a
(1)𝑎P𝑏*'}
(2)3𝑎+ 𝑏*
ëì(1)𝑎= 4a𝑏= −8
(2) ± 2
OíîĚěïðĜ!a*]9ač!a*]9$ª–p
_1`&!a*]9–ę2𝑎+ 1 = 9a3𝑎+ 2𝑏−4 = −8aºÂª–„ aab *'p
_2`ŀ aab *'Ķķ3𝑎+ 𝑏aÁª–p
ö$_1`$
 ∵2𝑎+ 1*z ± 3a3𝑎+ 2𝑏−4*!z−2a
∴2𝑎+ 1 = 9a3𝑎+ 2𝑏−4 = −8a
∴𝑎= 4a
∴𝑏= −8p
_2`$
∵3𝑎+ 𝑏= 3 × 4−8 = 4a
∴3𝑎+ 2𝑏*‘ ± 2p
【题型4  实数的运算】
£ 4_24-25 ÞßÃu·ŁłŃń·äA`ĸ 1a 2a 3a 6•{Ņ2ņæa¤X](𝑚,𝑛)ŒŇ𝑚ņÄ
ÅĝÇŇ𝑛c)a¦(5,4)Q(12,4)Œ*…)†ňz         
.
ëì0
OíÑĚěïðĜ)ʼn*ý>XYa()*+,aŊ”)ʼný>XYzĆp
č)*ņæ˖ęaŇbņ
1 c)aŇ0ņ 2 c)pŇŋņ 3 c)aŇŁņ 4 c)a…Ň𝑚ņ7𝑚c
)aÄŇbņÆ𝑚ņŌ7
1 + 2 + 3 + 4 + … + 𝑚=
𝑚(𝑚+1)
2
c)ač)*ņæËa™Łc)bcōŎa
čŏ8ŐŊ„Ň𝑚ņŇ𝑛c)ÆőzŒc)ÂÁİ,p
ö$$
č)*ņæ˖ęaŇbņ
1 c)aŇ0ņ 2 c)pŇŋņ 3 c)aŇŁņ 4 c)a…
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Ň𝑚ņ7𝑚c)aÄŇbņÆ𝑚ņŌ7
1 + 2 + 3 + 4 + … + 𝑚=
𝑚(𝑚+1)
2
c)ač)*ņæËa™Łc
)bcōŎa
(5,4)ŒŇ 5 ņÄÅĝÇŇ 4 c)z 2a
∵œ 11 ņŌ71 + 2 + 3 + 4 + … + 11 =
11×(11+1)
2
= 66 (c))a
∴ (12,4)ŒŇ 12 ņŇ 4 c)ªŇ 70 c)a
∵70 ÷ 4 = 17⋯2a
∴ (12,4)Œ*)z 2a
∴ (5,4)Q(12,4)Œ*…)†ňz 2−2 = 0a
Öëì‘
0p
ý2 4-1_2025 ŔßÛ·Ďď·ŕĄą`İ,
 16−(−1)2025−
3 27−|1−2|p
ëì3−2
OíîĚěïðĜ()*+,aÀİ,,êP!aŖİ,© 'P´a@Âİ,²³
˪–ġÆëìp
ö$$
ŗ2 = 4−(−1)−3−( 2−1)
= 4 + 1−3−2 + 1
= 3−2p
ý2 4-2_24-25 ŔßÛ·ħāŘń·äå`X]
[𝑚]ŒmřŸ𝑚*@M)a{𝑚}Œ𝑚*J)NOa
[𝑚] + {𝑚} = 𝑚a6A𝑚‘()p£{[3.14] = 3a{3.14} = 0.14a[3.14] + {3.14} = 3.14pİ,
[−3]−{ 5}
=       
p
ëì−5
OíîïðĜ\]9u*()+,a-.)*/,a.$\]9z$*ĆpOi/İ−3P 5
«Œ…cM)†‡aÁč\]9ġ„[−3] = −2a{ 5} =
5−2a…Śy³ª–ġ„ëìp
ö$$
∵−2 < −3 < −1a2 <
5 < 3a
∴[−3] = −2a{ 5} =
5−2a
∴[−3]−{ 5} = −2−( 5−2) = −5p
Öëì‘
−5p
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ý2 4-3_24-25 ÞßÛ·śľŜÿ·äA`™c%¾ă&¤ŝ:ŞşŠŽašŢİĜbă+,%¾{
Ņ
£{
Ùţķ x *'‘−1Ûaİ,Üĉ 3 < 4}ĸţķ'ý‘(−1) +1Ť0aİ,Üĉ‘ 6 < 4}Áĸţķ
'ý‘Ĝ0 + 1 = 1aťŦ+,aÆİ,ÜĉmJÍ 4aŧţ„ŨÜĉp
ũŐïuæĦp
(1)Ùţķ x *'‘ 5a¦ţ„ y *'z‰Ūūũæ2İ,p
(2)ÙŬŭţķ x *'‘ 1aũԟİ,çŮůŸv1%¾+±Âŧ‹ţ„ yp
ëì(1) 21
(2)4 1
OíîïðĜ()*+,a.$8aćĈŰŅA*+,˦z$*Ćp
_1`čŰŅA*+,%¾İ,ª–}
_2`čŰŅA*+,%¾İ,aÆÜĉÍC|Í 4 ªţ„Üĉ‘űp
ö$_1`Ùţķ x *'‘ 5 Ûa
¦75−(−2) = 7a7 × 3 = 21a
¸ 21 > 4a
∴ţ„ y *'z 21p
_2`Ùţķ x *'‘ 1 Ûa
¦71−(−2) = 3a3 × 3 = 9a 9 < 4aťŦİ,}
Ň01ţķ x *'‘1 + 1 = 2Ûa
¦72−(−2) = 4a4 × 3 = 12a 12 < 4aťŦİ,}
Ňŋ1ţķ x *'‘2 + 1 = 3Ûa
¦73−(−2) = 5a5 × 3 = 15a 15 < 4aťŦİ,}
ŇŁ1ţķ x *'‘3 + 1 = 4Ûa
¦74−(−2) = 6a6 × 3 = 18a 18 > 4aţ„𝑦}
ĽtůŸ 4 1%¾+±Âŧ‹ţ„ yp
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【题型5  无理数的估算】
£ 5_24-25 ÞßÃu·ŲųŴŵ·äå`Ŷºzŷ*ŢİaªŸzbŹJJ*źŻażŽ‚ž“ŷ
”p{Ņa
𝐵𝐶
𝐴𝐵*'×ſƀK
5−1
2 a¦uæ/,—Õ*z_    `
Ap0 <
5−1
2
<
2
5
Bp
2
5 <
5−1
2
<
1
2
Cp
1
2 <
5−1
2
< 1
Dp1 <
5−1
2
< 2
ëìC
OíîïðĜ-.)*/,aîïð-.)*/İam|2*<=a—ÕƁƂ*¬­z$î
*Ćp&8ę4 < 5 < 9a¦2 <
5 < 3ačm|2*<=İ,$ª–p
ö$$
&8ęa 4 < 5 < 9a
∴2 <
5 < 3a
∴1 <
5−1 < 2a
∴
1
2 <
5−1
2
< 1p
Öô
Cp
ý2 5-1_24-25 ŔßÃu·ľƃƄƅ·Ă㥹`Ęę432 = 1849a442 = 1936a452 = 2025a462
= 2116p¤𝑛‘M)¸𝑛<
2022 < 𝑛+ 1a¦𝑛*'‘_   `
Ap43
Bp44
Cp45
Dp46
ëìB
OíîĚěïð-.)*/,aĠĄćĈ-.)*/,z$*Ć}ÀčŝA*)č/, 2022
*Ja°·Ħ–$p
ö$$
&8–ę
1936 < 2022 < 2025
∴44 <
2022 < 45a
∴𝑛= 44}
Öô Bp
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ý2 5-2_24-25 ĢßÛ·ľƃƄƅ·äA`¤𝑎<
23 < 𝑏a¸𝑎 𝑏‘ƆŦ—M)a¦𝑎+ 𝑏=        
ëì9
Oíîïð()*/,QJKL*‹ƇaÀ/,„ 23*ƈ'¬­aġ„𝑎a𝑏*'a°·–ġ„Ü
Ýpč8„𝑎a𝑏*'z$*Ćp
ö$$
∵16 < 23 < 25a
∴4 <
23 < 5a
∵𝑎a𝑏‘…cƆŦM)a
∴𝑎= 4a𝑏= 5a
∴𝑎+ 𝑏= 4 + 5 = 9p
Öëì‘
9p
ý2 5-3_24-25 ÞßÃu·ĬƉà·äå`Ęę𝑘<
17−1a¸𝑘‘—M)a¦𝑘*'–tz           
_ƒ„bcª–`p
ëì1_ëìmƊb`
OíîĚěïðĜ-.)*/, m|2*$|ęļœaćĈ-.)*/,Ëz$*Ćp
À/, 17ġƈ'¬­aÁÕ] k *ƈ'¬­aºÂčm|2$*]9ª–$ëp
ö$$
∵16 <
17 <
25zM)}
∴3 <
17−1 < 4zƋĎ)}
∴𝑘≤3a
∵𝑘‘—M)a
∴𝑘= 1C 2 C 3p
Öëì‘
1_ëìmƊb`p
【题型6  二次根式的概念及其有意义的条件】
£ 6_24-25 ŔßÃu·àáƌÿ·Ă㥹`Ęę 3𝑥−6 +
6−3𝑥+𝑦= 2025a¦ 2025𝑥𝑦*'‘_`
Ap2025 3
Bp2025 2
Cp2024
Dp2025
ëìB
OíÀč012789*:;„𝑥*'aÁĶķ„𝑦*'a@Âİ, 2025𝑥𝑦pîĚěïð
Ĝ012789*:;aĠĄćĈ012Agf)zHI)z$*Ćp
ö$$
&8ġ
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3𝑥−6 ≥0
6−3𝑥≥0
 
a
$3𝑥−6 ≥0ġ𝑥≥2}
$6−3𝑥≥0ġ𝑥≤2p
∴𝑥= 2p
ŀ𝑥= 2Ķķ 3𝑥−6 +
6−3𝑥+𝑦= 2025ġ
3 × 2−6 +
6−3 × 2 +𝑦= 2025a
$ġ𝑦= 2025p
∴2025𝑥𝑦=
2025 × 2 × 2025 =
20252 × 2 = 2025 2p
Öô
Bp
ý2 6-1_24-25 ŔßÃu·Ĺááƍ·Ă㥹`uæó2Aab]z012*z_ƎƎ`
Ap −2
Bp
3 3
Cp 5
Dp 𝑎+ 1
ëìC
OíîïðĜ012*]9as{ 𝑎(𝑎≥0)*2Əd012ač]9òõOíª–p
ö$$
Ap∵−2 < 0a∴−2mz012aÖÑôõmjT8}    
Bp∵
3 3*Ş)z 3a∴
3 3mz012aÖÑôõmjT8}    
Cp∵5 > 0a∴5z012aÖÑôõjT8}
DpÙ𝑎+ 1 < 0ª𝑎< −1Ûa 𝑎+ 1mz012aÖÑôõmjT8}    
Öô
Cp
ý2 6-2_24-25 ŔßÃu·ħāƐń·fïƑ`Ęę
𝑦=
𝑥−4+ 4−𝑥
3
−2a¦3𝑥+ 4𝑦=      
p
ëì4
OíÑĚěïðĜ012789*:;a—Õ$m|2Šz$Ćp
"#012789*:;ġ„m|2Š„ xay *'aºÂĶķİ,p
ö$$
∵𝑦=
𝑥−4+ 4−𝑥
3
−2a
∴
𝑥−4 ≥0
4−𝑥≥0
 
a
$ġ
𝑥= 4a
Ö𝑦= −2a
¦3𝑥+ 4𝑦= 3 × 4 + 4 × (−2) = 4p
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Öëì‘
4p
ý2 6-3_24-25 ŔßÃu·ħāƐń·fïƑ`¤𝑥 𝑦‘()a¸÷ø2 4−𝑥+ |𝑦| = 𝑥−4a¦ 𝑥−𝑦*
,ê‘      
p
ëì 2
Oíîïð012789*:; HI)*<= ,êaÀč,êP© '*
HI<aÜT012789*:;ġ x y 'a°·ġ 𝑥−𝑦ª–$p
ö$$
∵2 4−𝑥+ |𝑦| = 𝑥−4a2 4−𝑥≥0a|𝑦| ≥0a
∴4−𝑥≥0¸𝑥−4 ≥0a
∴4−𝑥= 𝑥−4 = 0a$ġ𝑥= 4a
ĸ𝑥= 4Ķķ2 4−𝑥+ |𝑦| = 𝑥−4Aaġ𝑦= 0a
∴𝑥−𝑦=
4−0 =
4 = 2a
∴𝑥−𝑦*,ê‘ 2p
Öëì‘
 2p
【题型7 利用二次根式的性质化简】
£ 7_24-25 ŔßÃu·ŲųŴŵ·Ă㥹`ŀO2𝑎−
1
𝑎a	Ò*ʼnW a Ğ°	®*Üĉz_   `
Ap 𝑎
Bp −𝑎
Cp−𝑎
Dp−−𝑎
ëìD
OíĚěïðĜ012*89p$*Ćz‹—Õ*ŀ	Ò*Ķ)2C)ʼnĞÆ	®NaĪ
zf*ƒ+,aÄ	ÒĞÆ	®ěaŖ¸ĞÆ	®Qŗ	®*2Əz´Ɠ*Ìp8
	Ò*)ʼnC2ƏzI)ÛaĶŒMc2ƏzI'aěŀI	ƔÆ	ÒÁÂĞÆ	®p
{ĉ	Ò*)ʼnC2ƏzI)ÛaĶŒMc2ƏzI'aěŀI	ƔÆ	ÒÁÂĞÆ	®aº
Â>?ª–p
ö$$
&012*89–ę𝑎< 0a
∴𝑎−
1
𝑎= −−
1
𝑎× (−𝑎)2 = −−𝑎aÖ D —Õp
Öô
Dp
ý2 7-1_24-25 ĢßÛ·Łłƕà·Ă㥹`¤𝑥𝑦> 0a¦ −𝑥2𝑦>?‘_  `
Ap𝑥−𝑦
Bp−𝑥𝑦
Cp−𝑥𝑦
Dp−𝑥−𝑦
ëìD
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OíîïðĜ012*>?aĠĄćĈ012*<=z$ĆpÀƁƂ„𝑥< 0a𝑦< 0aÁ"
#012*<=>?ª–ġp
ö$$
∵𝑥𝑦> 0a
∴𝑥,𝑦¿	a¸ùm‘ 0a
Ɩ∵« −𝑥2𝑦Aa−𝑥2𝑦zgf)a
∴−𝑥2𝑦> 0a
∴𝑦< 0a
∴𝑥< 0a
∴−𝑥2𝑦= |𝑥| −𝑦= −𝑥−𝑦a
Öô
Dp
ý2 7-2¤𝑚< 0a¦ŀ2Ə𝑚𝑥	ÒÊ*ü2ĞÆ	ÉÂ*2ƏUz_ `
Ap−𝑚2𝑥
Bp −𝑚2𝑥
Cp (−𝑚)2𝑥
Dp−−𝑚𝑥
ëìA
Oíč m<0a¦012MƗzI)a¦ m ĞÆ	®NÂaÒÊ7I	a°·ġ„ëì.
ö$$
∵m<0,
∴ŀ2Ə𝑚𝑥	ÒÊ*ü2ĞÆ	ÉÂ*2Ə‘
−𝑚2𝑥.
Öô A.
œĥÑĚěïðĜ012*>?'ač012MƗ*j	ġ„z$*Ć.
ý2 7-3¤ 1575𝑛zM)aƒ„bc—M)𝑛*–‹'      
p
ëì7_ëìmƊb`
OíěƘƙ1575𝑛‹f‘M)aĸ1575ü)O$a°·ª–$p
ö$$
∵1575 = 9 × 25 × 7 = 32 × 52 × 7a
¤ 1575𝑛zM)a¦𝑛*'–tz7
Öëì‘
7_ëìmƊb`p
œĥÑĚěïðĜ012*<=a—Õ>?012ġ„z$Ćp
【题型8  最简二次根式】
£ 8_24-25 ŔßÛ·ƚĺƛā·Ă㥹`Ęę 𝑥2 + 1z@?012a¸Q 5–tTŖa¦𝑥*'
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‘_    `
Ap2
Bp−4
Cp4
Dp ± 2
ëìD
OíîïðĜ¿’012a@?012a"#$%aĠĄćĈęļœ*U#z$*
Ćp
& 𝑥2 + 1z@?012¸Q 5–tTŖaġ„𝑥2 +1 = 5aºÂ"#$%ª–p
ö$$
∵𝑥2 + 1z@?012¸Q 5–tTŖa
∴𝑥2 +1 = 5a$ġ
𝑥=± 2a
Öô
Dp
ý2 8-1_24-25 ŔßÛ·›ƜſƝ·äå`uæ012Aaz@?012*z_   `
Ap 18
Bp 63𝑎
Cp 𝑎2−2𝑎+ 4
Dp
𝑎
3
ëìC
OíîïðĜ@?012*]9aƁ]bc012zmz@?012*ËaƞzòcƟð
@?012*…c:;zƠ¿Û÷øa¿Û÷ø*ƞz@?012aƠ¦ƞmzp
÷øuæ…c:;*012ade@?012
_1`gf)*ü)zM)aü2zM2}
_2`gf)Am‚‹fġx*ü)Cü2p
–tѝƁƂŒcôõz—Õ*p
ö$$
A 18=3 2amz@?012aÖîôõmjT8}
B 63𝑎= 3 7𝑎amz@?012aÖîôõmjT8}
C 𝑎2−2𝑎+ 4z@?012aÖîôõjT8}
D 
𝑎
3 =
3𝑎
3 amz@?012aÖîôõmjT8}
Öô
Cp
ý2 8-2ĸ2Ə 35−𝑎_a ‘—M)`>‘@?012Âa–tQ 8TŖpƒ„bcjT:; a *
'      
p
ëì3_ëìmƊb`
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OíîïðĜ¿’012a@?012aĠĄćĈ¿’012*]9z$*Ćp
ö$$
∵8 = 2 2a
∴35−𝑎–t‘ 2a2 2a3 2a4 2a
∴35−𝑎= 2C35−𝑎= 8C35−𝑎= 18C35−𝑎= 32a
$ġ
𝑎= 33C𝑎= 27C𝑎= 17C𝑎= 3a
Öëì‘
3p
ý2 8-3Ęę𝐴= 2 2𝑥+ 1a𝐵= 3 𝑥+ 3a𝐶=
10𝑥+ 3𝑦a6A AaB ‘@?012a¸
𝐴+ 𝐵= 𝐶a¦2𝑦−𝑥*'‘      
p
ëì68
Oíč8ġ„2𝑥+ 1 = 𝑥+ 3a„𝑥= 2a°·ġ„10𝑥+ 3𝑦= (5 5)
2
= 125a„𝑦= 35aÁ
Ķķ'ª–p
ö$∵AaB ‘@?012a¸𝐴+ 𝐵= 𝐶a
∴2𝑥+ 1 = 𝑥+ 3a
$ġ𝑥= 2a
∴𝐴= 2 5a𝐵= 3 5a𝐴+ 𝐵= 5 5 = 𝐶a
∴10𝑥+ 3𝑦= (5 5)
2
= 125a
$ġ𝑦= 35a 
∴2𝑦−𝑥= 2 × 35−2 = 68p
Öëì‘
68p
œĥîïð@?012ač@?012*]9ġ„x = 2z$*Ćp
【题型9  求二次根式中的参数值】
£ 9¤(2−7)
2
= 𝑎−𝑏7a6A𝑎a𝑏ù‘M)a¦𝑏−𝑎*'‘      
p
ëì−7
Oíîïð012´Ë+,QĶ)2'aơ5ƋĎň25012y|a"#ƋĎ
ň2ƢfÂa&012y|*:;ġÆ𝑎a𝑏aĶķŒƣ2$ª–ġÆëìaĠĄćĈ012+,
z$ƤĦ*Ćp
ö$$
 ∵(2−7)
2
= 22−2 × 2 ×
7 + ( 7)
2
= 11−4 7 = 𝑎−𝑏7a
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∴𝑎= 11,𝑏= 4a
∴𝑏−𝑎= 4−11 = −7a
Öëì‘
−7p
ý2 9-1012 2𝑏+ 1Q 𝑎−1 *P‘ 0a¦𝑎+ 𝑏*'‘     p
ëì
1
2/0.5
OíîïðĜ012*HI<aM2*'}–ġ 2𝑏+ 1 +
𝑎−1 = 0a&012*HI<ġ
𝑎−1 = 0a2𝑏+ 1 = 0a„𝑎P𝑏aĶ'ª–$}.$012*HI< 𝑎≥0_𝑎≥0`z$*Ćp
ö$$
&8ġ
2𝑏+ 1 +
𝑎−1 = 0a
∴ 𝑎−1 = 0a2𝑏+ 1 = 0a
$ġ
𝑎= 1a𝑏= −
1
2a
∴ 𝑎+ 𝑏
= 1−1
2
=
1
2}
Öëì
1
2p
ý2 9-2_2025 ĢßÃu·Ĭ·ŕĄą`¤
𝑚
2 = 2 2a¦M)𝑚*'‘           
p
ëì16
Oíîïð012*+,a…Ï„ m *'ª–$p
ö$$
∵
𝑚
2 = 2 2a
∴
𝑚
2 = 8a
$ġ𝑚= 16a
Öëì‘
16p
ý2 9-3Ęę 10−𝑛zM)a¦—M)𝑛Ľ7–‹'Ō7      
caĪī*P‘      
p
ëì     
4     
26
Oíîïð012789*:; 012*<=ač012*gf)zHI)ÜT01
2*<=ġƗM) n *'a°·$ª–p
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ö$$
« 10−𝑛Aa10−𝑛≥0a¦𝑛≤10a
∵10−𝑛zM)a¸ n ‘—M)aƖ 0 = 0a 1 = 1a 4 = 2a 9 = 3a 
∴n *'‘ 10 C 9 C 6 C 1aŌ 4 ca
Īī*P‘10 + 9 + 6 + 1 = 26a
Öëì‘
4a26p
【题型10  二次根式的运算或化简求值】
£ 10Ęę𝑥=
2 +1a𝑦=
2−1auæó2*'p
(1)𝑥2 + 𝑦2}
(2)
𝑥
𝑦+
𝑦
𝑥p
ëì(1)6
(2)2 2
Oíîïð012*>?'aĠĄćĈ012*¼T+,˦z$*Ćp
_1`"#2𝑎2 + 𝑏2 = (𝑎+ 𝑏)2−2𝑎𝑏aĶķ𝑥=
2 +1a𝑦=
2−1aª–'}
_2`Àġ„(
𝑥
𝑦+
𝑦
𝑥)
2
=
𝑥
𝑦+2
𝑥
𝑦⋅
𝑦
𝑥+
𝑦
𝑥=
𝑥
𝑦+
𝑦
𝑥+2 =
𝑥2+𝑦2
𝑥𝑦
+2aÜT_1`Ķķ'ª–ġëìp
ö$_1`$
č𝑥=
2 +1,𝑦=
2−1a¦7
𝑥+ 𝑦= ( 2 +1) + ( 2−1) = 2 2a
𝑥𝑦= ( 2 +1)( 2−1) = 2−1 = 1a
Ķķ𝑥2 + 𝑦2ġÆ
𝑥2 + 𝑦2 = (𝑥+ 𝑦)2−2𝑥𝑦= (2 2)
2
−2 × 1 = 8−2 = 6}
_2`(
𝑥
𝑦+
𝑦
𝑥)
2
=
𝑥
𝑦+2
𝑥
𝑦⋅
𝑦
𝑥+
𝑦
𝑥=
𝑥
𝑦+
𝑦
𝑥+2 =
𝑥2+𝑦2
𝑥𝑦
+2a
ü‘𝑥2 + 𝑦2 = 6a𝑥𝑦= ( 2 +1)( 2−1) = 2−1 = 1a
Ľt(
𝑥
𝑦+
𝑦
𝑥)
2
=
𝑥2+𝑦2
𝑥𝑦
+2 =
6
1 +2 = 8a
Ľt
𝑥
𝑦+
𝑦
𝑥=
8 = 2 2p
ý2 10-1İ,
(1) 3 +3
1
3−12}
(2) 45 ÷
27 ×
3
5}
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(3) 18 ÷
2−32 +
3 ×
6}
(4)(3 +
3)(3−3) + (1 +
7)
2
p
ëì(1)0
(2)1
(3)3−2
(4)14 + 2 7
OíÑïðĜ012*¼T+,aĠĄćĈ012*+,˦zĆp
_1`ŀ012>?°±²³+,ª–}
_2`ÀŀµËƥ>‘´ËaÁ°±İ,ª–}
_3`Àİ,´µËaÁİ,²³ª–}
_4`"#ň2PƋĎ2°±İ,ª–p
ö$_1`$
 3 +3
1
3−12
=
3 +
3−2 3
= 0
_2` 45 ÷
27 ×
3
5
=
45
27 ×
3
5
=
45
27 × 3
5
=
1
= 1
_3` 18 ÷
2−32 +
3 ×
6
=
9−4 2 +
18
= 3−4 2 + 3 2
= 3−2
_4`(3 +
3)(3−3) + (1 +
7)
2
= 32−( 3)2 + (1 + 2 7 + 7)
= 9−3 + 1 + 2 7 + 7
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
= 14 + 2 7
ý2 10-2_24-25 ŔßÛ·Łłƣÿ·Ă㥹`Ęę𝑥= 2−3aĶ)2(7 +
3)𝑥2 + (2 +
3)𝑥+
3
*'p
ëì38−20 3
Oíîïð*zĘę:;2a$Ķ)2*'aÀ$𝑥2 = (2−3)
2
= 7−4 3aÁĶķ°±İ,ª
–p
ö$$
∵𝑥= 2−3a
∴𝑥2 = (2−3)
2
= 7−4 3a
∴(7 +
3)𝑥2 + (2 +
3)𝑥+
3
= (7 +
3) (7−4 3) + (2 +
3) (2−3) +
3
= 49−21 3−12 + 4−3 +
3
= 38−20 3p
ý2 10-3Ęę 𝑥+ 𝑦= −5a𝑥𝑦= 4a𝑥
𝑦
𝑥+𝑦
𝑥
𝑦*'p
ëì−4
OíÀč𝑥+ 𝑦= −5a𝑥𝑦= 4a–ƁƂ𝑥< 0a𝑦< 0aÁĸŗ2>?aºÂĸĘę:;MƗĶķ'
ª–p
îĚěïðĜ012>?5012*<=aĠĄćĈ 𝑎2 = |𝑎| =
𝑎(𝑎> 0)
0(𝑎= 0)
−𝑎(𝑎< 0)
 
z$*Ćp
ö$ ∵𝑥+ 𝑦= −5 < 0a𝑥𝑦= 4 > 0a
∴𝑥< 0a𝑦< 0a
∴ŗ2=𝑥
𝑥𝑦
𝑥2 +𝑦
𝑥𝑦
𝑦2 = −𝑥⋅
𝑥𝑦
𝑥−𝑦⋅
𝑥𝑦
𝑦= −2 𝑥𝑦 .
∵𝑥𝑦= 4p
ŗ2 = −2 ×
4 = −2 × 2 = −4p
【拔尖篇】
【题型11  二次根式的双重非负性】
£ 11_24-25 ÞßÃu·ħƦƧ·Ă㥹`Ęę aab ‘()a¸𝑏=
𝑎−8−8−𝑎+25a¦
3 𝑎+
𝑏
*'‘_    `p
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Ap10
Bp9
Cp8
Dp7
ëìD
Oíîïð,êŽ!*:;a!ač,ê*f)zHI)ġÆ𝑎= 8a°·
ġ𝑏= 25a@ÂĶķ
3 𝑎+
𝑏İ,ª–p
ö$$
∵𝑏=
𝑎−8−8−𝑎+25a𝑎−8 ≥0a8−𝑎≥0a
∴𝑎= 8a𝑏= 25a
∴
3 𝑎+
𝑏=
3 8 +
25 = 2 + 5 = 7a
Öô
Dp
ý2 11-1_25-26 ŔßÛ·Ďď·ƨƩĄą`¤𝑎+
𝑎−3 = 3a 𝑎+ 3*'p
ëì 6
OíîĚěïð*z012aĠĄćĈ012*<=z$*Ćp
Àĸŗ2>‘ 𝑎−3 = 3−𝑎aÁ#m|2Š$ª–p
ö$$
ü‘𝑎+
𝑎−3 = 3a
Ľt 𝑎−3 = 3−𝑎a
Ľt𝑎−3 ≥0,3−𝑎≥0a
Ľt𝑎= 3a
Ľt 𝑎+ 3 =
3 + 3 =
6p
ý2 11-2_24-25 ÞßÃu·þÿþń·äA`Ęę|𝑎−3| +
3 + 𝑏= 0a¦
𝑎
𝑏
2025
= _    `
Ap1
Bp−1
Cp2
Dp−2
ëìB
Oíîïð© 'a,êa7.)*´a$*Ćz„𝑎P𝑏*'p
č© 'P,ê*HI<a$ġ𝑎P𝑏*'aĶķİ,ª–p
ö$$
|𝑎−3| +
3 + 𝑏= 0a|𝑎−3| ≥0a 3 + 𝑏≥0a
∴|𝑎−3| = 0a 3 + 𝑏= 0a
∴𝑎−3 = 0a3 + 𝑏= 0a
∴𝑎= 3a𝑏= −3a
∴
𝑎
𝑏= −1
∴
𝑎
𝑏
2025
= (−1)2025 = −1
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Öô
Bp
ý2 11-3_2025 ŔßÃu·ħ·ŕĄą`ĘęH¶() aab ÷ø|2a−4| + |b + 2| +
(a−4)b2
+4 = 2𝑎a¦𝑎+ 𝑏=      
p
ëì2
OíîĚěïðĜ,ê*HI<a© '*HI<aĸ2Əýs‘|2𝑎−4| + |𝑏+ 2| +
(𝑎−4)𝑏2
= 2𝑎−4a&,ê*HI<a© '*HI<ġ„2𝑎−4 ≥0aÁ>?2Ə–ġ„|𝑏+ 2| +
(𝑎−4)𝑏2
= 0aÁč,ê*HI<a© '*HI<ª–ġ„𝑏= −2a𝑎= 4a°·ĶķĶ)2ª–ġ„ë
ìp
ö$$
∵|2𝑎−4| + |𝑏+ 2| +
(𝑎−4)𝑏2 +4 = 2𝑎a
∴|2𝑎−4| + |𝑏+ 2| +
(𝑎−4)𝑏2 = 2𝑎−4a
∴2𝑎−4 ≥0a
∴2𝑎−4 + |𝑏+ 2| +
(𝑎−4)𝑏2 = 2𝑎−4a
∴|𝑏+ 2| +
(𝑎−4)𝑏2 = 0
∴𝑏+ 2 = 0a𝑎−4 = 0a
$ġa𝑏= −2a𝑎= 4a
¦𝑎+ 𝑏= 4−2 = 2a
Öëì‘
2p
【题型12  实数的大小比较】
£ 12_24-25 ÞßÃu·ıIJƪƜ\·äA`KLJ
 11     3}−5     −2}
5+1
2     
3
2p_ē“ > ” 
“ = ”C“ < ”`p
ëì     
>     
<     
>
OíîïðĜ()*JKLaĠĄćĈt›ęļœz$ëî*Ćp
č()*JKLË$몖p
ö$$
 ∵ 
11 >
9a
∴
11 > 3}
∵5 >
4 = 2a
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∴−5 < −4a
∴−5 < −2}
∵5 >
4 = 2a
∴
5 +1 > 3a
∴
5+1
2
>
3
2}
Öëì‘
 > a < a > p
ý2 12-1_24-25 ÞßÃu·àáƌÿ·äA`¤𝑎=
3 7a𝑏=
5a𝑐= 2a¦ aabac *J̑
_    `
Ap𝑏≤𝑐< 𝑎
Bp𝑏< 𝑎< 𝑐
Cp𝑎< 𝑐< 𝑏
Dp𝑎< 𝑏< 𝑐
ëìC
Oíîïð()*JKLaÀ"#ƫƬË/, aab *'aÁKLJª–p
ö$$
∵1 < 7 < 8a4 < 5 < 9a
∴
3 1 <
3 7 <
3 8a 4 <
5 <
9a
ª1 <
3 7 < 2a2 <
5 < 3a
∴1 < 𝑎< 2a2 < 𝑏< 3a
Ɩ𝑐= 2a
∴𝑎< 𝑐< 𝑏a
Öô
Cp
ý2 12-2_25-26 ŔßÛ·Ďď·ƨƩĄą`KL 65−2Q 10 +2*Jp
ëì 65−2 >
10 +2
OíÑïð-.)*/,P()*JKLa–ÔŸ/,…c-.)*¬­a°·Õ]…cĶ)2
*¬­aÁ°±KLp
ö$$
ü‘8 <
65 < 9a3 <
10 < 4a
Ľt6 <
65−2 < 7a5 <
10 +2 < 6a
Ľt 65−2 >
10 +2p
ý2 12-3_25-26 ŔßÛ·Ďď·ƨƩĄą`KLuæóŠ)*J
(1)
1
2Q
10
6 }
(2)
6−1
2 Q 1p
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ëì(1)
10
6 >
1
2
(2)
6−1
2
< 1
Oíîïð()KLJaĠĄćĈ()KLJ*Ëaz$*Ć
_1`/,ËKLJª–}
_2`/,ËKLJª–p
ö$_1`$
∵10 >
9 = 3a
∴
10
6 >
3
6aª
10
6 >
1
2}
_2`∵4 <
6 <
9a
∴2 <
6 < 3a
∴1 <
6−1 < 2a
∴
6−1
2
<
2
2aª
6−1
2
< 1p
【题型13  求整数部分和小数部分】
£ 13_24-25 ŔßÛ·Ĭā\ƭ·Ă㥹`¤ 3*M)NO‘ aaJ)NO‘ ba4−3*M)NO‘
caJ)NO‘ da¦
𝑏+𝑑
𝑎𝑐*'‘_   `
Ap
1
2
Bp
1
4
Cp
3−1
2
Dp
3+1
2
ëìA
OíîïðĜ/,-.)*Jaԟ/, 3«Œ…cM)†‡aÄ·Õ]𝑎= 1a𝑏=
3−1aÔ
Ÿ/,4−3«Œ…cM)†‡aÄ·Õ]𝑐= 2a𝑑= 2−3aºÂŀ a b c d Ķķġ)'p
ö$$
∵3*M)NO‘ aaJ)NO‘ ba1 <
3 < 2a
∴𝑎= 1a𝑏=
3−1a
∵1 <
3 < 2a
∴−2 < −3 < −1a
∴2 < 4−3 < 3a
∵4−3*M)NO‘ caJ)NO‘ da
∴𝑐= 2a𝑑= 4−3−2 = 2−3a
∴
𝑏+𝑑
𝑎𝑐=
3−1+2−3
1×2
=
1
2p
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Öô
Ap
ý2 13-1_24-25 ÞßÃu·ĹáĹÿ·äA`¤6 +
5*M)NOz𝑚aJ)NOz𝑛a¦|𝑛−𝑚|‘_   `
Ap 5−10
Bp10−5
Cp 5−6
Dp6−5
ëìB
OíîïðQ-.)*M)NO7*İ,a()*+,aƫƬË„𝑚,𝑛*'aÁĶ'İ,ª–p
ö$$
∵4 <
5 <
9a
∴2 <
5 < 3a
∴8 < 6 +
5 < 9a
∴𝑚= 8a𝑛= 6 +
5−8 =
5−2a
∴|𝑛−𝑚| = | 5−2−8| = 10−5}
Öô Bp
ý2 13-2_24-25 ŔßÛ·ħāŘń·äå`X]
[𝑚]ŒmřŸ𝑚*@M)a{𝑚}Œ𝑚*J)N
Oa[𝑚] + {𝑚} = 𝑚a6A𝑚‘()p£{[3.14] = 3a{3.14} = 0.14a[3.14] + {3.14} = 3.14pİ,
[−3]−{ 5} =       
p
ëì−5
OíîïðĜ\]9u*()+,a-.)*/,a.$\]9z$*ĆpOi/İ−3P 5
«Œ…cM)†‡aÁč\]9ġ„[−3] = −2a{ 5} =
5−2a…Śy³ª–ġ„ëìp
ö$$
∵−2 < −3 < −1a2 <
5 < 3a
∴[−3] = −2a{ 5} =
5−2a
∴[−3]−{ 5} = −2−( 5−2) = −5p
Öëì‘
−5p
ý2 13-3_24-25 ÞßÃu·ĹáľƮ·Ă㥹`Ęę()8 +
13*M)NO‘𝑎aJ)NOz𝑚}(
)9−13*M)NO‘𝑏aJ)NOz𝑛p
(1)ƒ„𝑎a𝑚a𝑏a𝑛*'}
(2)12𝑚+ 12𝑛+
𝑎+ 𝑏−7*'*}
(3)
3 (𝑚+ 𝑛)2025 +(𝑎−2𝑏)+ 6*'p
ëì(1)𝑎= 11a𝑚=
13−3a𝑏= 5a𝑛= 4−13
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(2) ± 3
(3)2
OíîïðĜ-.)*/,  !aĠĄćĈt›ęļœŖƯư+#z$Ñ*Ćp
_1`/,ġ„3 <
13 < 4aÄ·–ġ11 < 8 +
13 < 12a5 < 9−13 < 6aÜT8ª–ġ$}
_2`ĸ_1`A𝑎= 11a𝑚=
13−3a𝑏= 5a𝑛= 4−13ĶķĽ2Ə°±İ,aÁÜT*]9İ
,ª–ġ$}
_3`ĸ_1`A𝑎= 11a𝑚=
13−3a𝑏= 5a𝑛= 4−13ĶķĽ2Ə°±İ,aÁÜT!*]9İ
,ª–ġ$p
ö$_1`$
∵9 < 13 < 16a
∴9 <
13 <
16aª3 <
13 < 4a
∴11 < 8 +
13 < 12a5 < 9−13 < 6a
∵()8 +
13*M)NO‘𝑎aJ)NOz𝑚a()9−13*M)NO‘𝑏aJ)NOz𝑛a
∴𝑎= 11a𝑚= 8 +
13−11 =
13−3a𝑏= 5a𝑛= 9−13−5 = 4−13}
_2`$
&_1`–ġ
𝑎= 11a𝑚=
13−3a𝑏= 5a𝑛= 4−13a
12𝑚+ 12𝑛+
𝑎+ 𝑏−7
= 12 × ( 13−3) + 12 × (4−13) +
11 + 5−7
= 12 13−36 + 48−12 13 + 4−7
= 9a
∴12𝑚+ 12𝑛+
𝑎+ 𝑏−7*'*‘ ±
9 =± 3}
_3`$
3 (𝑚+ 𝑛)2025 +(𝑎−2𝑏)+ 6
=
3 ( 13−3 + 4−13)2025 + (11−2 × 5) + 6
=
3 1 + 1 + 6
=
3 8
= 2p
【题型14  实数与数轴综合运用】
£ 14_24-25 ÞßÃu·ĬƱƲƳ·äå`ƴĢƵ,êƶAƷƸêƹ
“ƷƸóŶ´aŖ·fµ†aª
ƺ”aª𝑐=
𝑎2 + 𝑏2_𝑎‘“Ʒ”a𝑏‘“Ƹ”a𝑐‘“ƺ”`¤“Ʒ”‘2a“Ƹ”‘3a¦“ƺ”«{ŅĽ)R›–Œ
«_    
`
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Ap𝐴œ
Bp𝐵œ
Cp𝐶œ
Dp𝐷œ
ëìC
Oíîïð-.)*/,a()Q)Ra.$8Ŗæġ—Õ*,2z$*Ćpč8æ2
İ,Â/,6JaºÂÕ]6«)R›*Šƻª–p
ö$$
¤“Ʒ”‘2a“Ƹ”‘3a¦ 22 + 32 =
13a
∵9 < 13 < 16a
∴3 <
13 < 4a
¦“ƺ”«{ŅĽ)R›–Œ«𝐶œa
Öô
Cp
ý2 14-1_24-25 ÞßÃu·Ƽãƽà·äA`{Ņa—s𝐴𝐵𝐶𝐷*ÊƓ‘ 7aƾœ A «)R›Œ*)
‘−1a¤œ E «)R›_œ E «œ A *Åƿ`a¸𝐴𝐷= 𝐴𝐸a¦œ E ĽŒ*)‘          
p
ëì−7−1
Oíîïð,êa()Q)RaÀ„—s*ÏĨa°·腜‡*ǀǁ„œ E ĽŒ
*)ª–p
ö$$
∵—s𝐴𝐵𝐶𝐷*ÊƓ‘ 7a
∴𝐴𝐸= 𝐴𝐷=
7a
∵ƾœ A «)R›Œ*)‘−1a
∴œ E ĽŒ*)‘−7−1}
Öëì‘
−7−1p
ý2 14-2_24-25 ŔßÃu·ǂŴľǃDŽľ·äå`()𝑎a𝑏«)R›*Šƻ{ŅĽa>?
|𝑎+ 1|−
(𝑏−1)2 +
(𝑎−𝑏)2 =           
p
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ëì2
OíîïðĜ()Q)Ra"#)Rġ„−1 < 𝑎< 0a1 < 𝑏< 2a°·>?ª–p
ö$$
&)Raġ−1 < 𝑎< 0a1 < 𝑏< 2a
∴𝑎+ 1 > 0a𝑏−1 > 0a𝑎−𝑏< 0a
∴ŗ2 = 𝑎+ 1−|𝑏−1| + |𝑎−𝑏|
= 𝑎+ 1−(𝑏−1)−(𝑎−𝑏)
= 𝑎+ 1−𝑏+ 1−𝑎+ 𝑏
= 2a
Öëì‘
2p
ý2 14-3_24-25 ÞßÃu·ħāqƌ·äA`{ŅaԟDžÏĨ‘ 1 *—saƞ‹”Õ*ŀ 2Œ«
)R›œ𝐴1džaLJ𝐴1Çƿ@×*M)œ‘𝐵1atœ𝐵1‘~Lja𝐴1𝐵1‘ljNJDžlj~aNj)R͜𝐴2aLJ𝐴2Ç
ƿ@×*M)œ‘𝐵2atœ𝐵2‘~Lja𝐴2𝐵2‘ljNJDžlj~aNj)R͜𝐴3a{ÑťŦa¦𝐴2026𝐵2026*Ĩ
‘          
p
ëì 2−1/−1 +
2
OíîïðĜ()*+,*XYa()Q)RaÀ„œ B Œ*)ġÆ𝐴1𝐵1 = 2−2a¦𝐴2Œ
*)‘4−2aÁ„𝐵2Œ*)‘3a¦𝐴2𝐵2 =
2−1aºÂˆ1Œ𝐴3𝐵3 = 2−2a𝐴4𝐵4 =
2−1a𝐴5𝐵5
= 2−2}𝐴6𝐵6 =
2−1}ª–ŊÆXY$p
ö$$
&8ġaœ𝐴1Œ*)‘ 2a
∵1 <
2 <
4a
∴1 <
2 < 2a
∴𝐵1Œ*)‘ 2a
a∴𝐴1𝐵1 = 2−2a¦𝐴2Œ*)‘2 + 2−2 = 4−2a
∵1 <
2 < 2a
∴−2 < −2 < −1
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
∴2 < 4−2 < 3a
∴𝐵2Œ*)‘3a
∴𝐴2𝐵2 = 3−(4−2) =
2−1a
¿.–ġ𝐴3𝐵3 = 2−2}
𝐴4𝐵4 =
2−1}
𝐴5𝐵5 = 2−2}
𝐴6𝐵6 =
2−1}
……a
tђnj–ġaÙ n ‘Ǎ)Û𝐴𝑛𝐵𝑛= 2−2aÙ n ‘ǎ)Û𝐴𝑛𝐵𝑛=
2−1a
∴𝐴2026𝐵2026 =
2−1
Öëì‘
 2−1p
【题型15  实数运算的应用】
£ 15_24-25 ÞßÃu·ħāĨĩ·Ă㥹`Ǐǐ7bcƗƓ‘200cm3*ǑǒŗǓaǏǐƘěĸČ
cǑǒǔ>ŖG\ǕǖŽ\*sǗp
(1)¤ĸŗǓG\ǕǖŽbcőʑ—s ǘ‘8cm*ĨƗaĨƗőʗs*ÏĨp
(2)ǏǐĖĸŗǓǕǖŽŋcJy¿*—ƗǑǒaǙđƋŽÂǚu*ĿǓƗƓ‘8cm3aǙđŽ*™
bcJ—ƗǑǒ*ǛĨp
ëì(1)5cm
(2)4cm
Oíîïð,êa!*U#a
_1`čĨƗƗƓ*İ,2“ĨƗ*ƗƓ = őÊƓ × ǘ”æ%$ª–}
_2`č“—ƗƗƓ*İ,Ët53cJ—ƗƗƓQÐƗƓ†‡*Ì”æ%$ª–}
.$,ê !*]9z—Õ$ë*Ćp
ö$_1`$
ŢĨƗőʗs*ÏĨ‘𝑥cma
ˆ8aġ
8𝑥2 = 200a
$ġ
𝑥= 5C𝑥= −5_I'mjT8aǜǝ`a
ë
ĨƗőʗs*ÏĨ‘5cm}
_2`$
ޙbcJ—ƗǑǒ*ǛĨ‘𝑦cma
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
ˆ8aġ
3𝑦3 +8 = 200a
$ġ
𝑦= 4a
ë
™bcJ—ƗǑǒ*ǛĨ‘4cmp
ý2 15-1_24-25 ÞßÛ·śľ·äA`{ŅabcőÊljNJ‘3cm*ǞƏ®ǟžbǠǡpÙǞƏ—
ǢÛaǞ®Ǡǡ*ǘǣ‘16cm}§ǢÛaėĿNO*ǘǣ‘4cmpǞ®*Ǡǡ—Ǥ§÷ 2 cbǥ*—
ƗǦǧ_πƈ 3aǦǧ*Ǩǣmİ`p
(1)ŨǞƏ*ǦƓ_ǟ÷ÛǠǡ*ƗƓ`z‰Ū!ǩǪū
(2)—ƗǦǧ*ǛĨz‰ŪǩǪū
ëì(1)540cm3
(2)6cm
Oíîïð7.)*¼T+, bc)*!aÓơ5qrv¨Ɨ*ƗƓaǫǬ8aġ„“Ǟ
Ə*ǦƓQ¿ő ǘ‘(16 + 4)cm*~ǭƗƓy|”z$*Ćp
_1`ǞƏ*ǦƓQ¿ő ǘ‘(16 + 4)cm*~ǭƗƓy|a&і$}
_2`ǮÀ„Ǟ®*Ǡǡ*ƗƓaºÂčǞ®*Ǡǡ—Ǥ§÷ 2 cbǥ*—ƗǦǧ$ª–p
ö$_1`$
𝑉= π𝑟2ℎ≈3 × 9 × (16 + 4) = 27 × 20 = 540(cm3)}
_2`$
ü‘𝑉溶液= π𝑟2ℎ≈3 × 32 × 16 = 432(cm3)p
ĽtǛĨ =
3 432 ÷ 2 = 6cmp
ý2 15-2ǯbǰDZ›ģÂa‘ĜŒDz«ĐäˆdzǴǵƋŽ)đĒ*JǶPJǷ¿a)šƤ]
«ǸÒǹǺ›œ 2 ǻǼǽšz 16 Ĥ*ǾǿȀȁȂīpÄǾǿįÆDZ*™ȃȄǼȅȆȇš‘ 1.6 Ĥa&
Í)šzŨǺ*ȈȉaüљǼš–tŸ#bcǺȊȆ* 5 ĤǺȋȌ ™ȃȄǼ*Ðǽ³ȍ 5 Ĥ
_ȄǼÐǽm‚ȅȆȇa¿bȄǼȎȏȐŸ#bcȋȌ`pȑčŨǾǿį*ȒưğaÙȄǼÐǽ_m‚
ȅȆȇ`÷ 30 ĤÛa5 Ĥ*ǺȋȌ–ȓȔ‘bc 7 Ĥ*įȕȋȌaª–tȖȄǼÐǽ_m‚ȅȆȇ`³
ȍ 7 ĤÙ)š¿ÛœĜ 2 ǻǾǿ”ȗuǼȘșÛaJǶ¿ç
“šaȚī–tȔbwuǼ2a
Ȓț‰Ȝ”ũ¿īOíJǶ¿*uǼ2aŖİ,„î1ÒǹÐȇ#_Ȍ‚ȅȆȇ`@ȝ–‘        
Ĥp
ëì25.2
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
OíOiİ,…wuǼ*2aKLŒbwÐȇ#țȝª–p
ö$ŇbwuǼ2‘Ȟȟ…ǻǾǿ
Tİȇ#‘
16 + 16 + 1.6−7 = 26.6Ĥ
Ň0wuǼ2‘u…cȄǼa™cȄǼȟbǻǾǿ
Tİȇ#‘
(16 + 1.6−5) × 2 = 25.2Ĥ
ÖôȠŇ0wțȡ,a@ȝȇ#‘ 25.2 Ĥ
Öëì‘
25.2p
œĥîïðĜ()+,*(^U#aO’ȢÝz$*Ćp
ý2 15-3ǸJh”ȗfĕbǒĨ‘32maȣ‘21m*ĨsėȤa
(1)ìb
{ŅaĸČǒėȤw›ȥȦaA‡ȧb:Ȩȩ*JȪaJȪ*ÅÏȫĝÇĞ1mƞzĪ*ÇÏ
ȫp¦ČǒȥȤ*ÊƓ‘ _____ m2}
(2)ì0
ȧųbcĨzȣ* 1.5 ȬaÊƓ‘384m2*ȭȮȯa¤KȰ#*ȭȮȯěĨ«23mÆ30m†
‡aȣ«13mÆ20m†‡p ČcȭȮȯ‹#eKȰȱū ŖçŮ.&p
ëì(1)651
(2)‹a.&r$í
OíîïðĜŅs*Ğa*]9|ęļp
_1`&8aȥȤ*Ĩ³J1maȣmýaü·–ġȥȤ*ÊƓ}
_2`Ţȣ𝑥ma¦Ĩ‘1.5𝑥mačÊƓ2ª–ġÍ x *%a&*]9ª–ġ xaÁ  x *
'°±/,a¤÷ø8ª–aƠ¦m±p
ö$_1`$
&8aJȪ*ÅÏȫĝÇĞ1mƞzĪ*ÇÏȫªJȪ*ȣ‘1ma
¦ȥȤ*Ĩ³J1maȣmýa
ÊƓ‘21 × (32−1) = 651m2}
Öëì‘
651p
_2`‹a.&{u
Ţȣ𝑥ma¦Ĩ‘1.5𝑥ma
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
ˆ87
1.5𝑥2 = 384a 𝑥2 = 256
∵𝑥> 0a
∴𝑥= 16a1.5𝑥= 24
jTĨ«23mÆ30m†‡aȣ«13mÆ20m†‡a
∴ČcȭȮȯ‹#eKȰp
【题型16  复合二次根式的化简】
£ 16Ȳȳuæó2
5 + 2 6 = (2 + 3) + 2 2 × 3 = ( 2)
2
+ ( 3)
2
+2 2 ×
3 = ( 2 +
3)
2
a
8 + 2 7 = (1 + 7) + 2 1 × 7 = 12 + ( 7)
2
+2 × 1 ×
7 = (1 +
7)
2
a……pũ+#t›*Ë>?
7 + 2 10 =        
p
ëì 5 +
2/ 2 +
5
OíîïðĜVT012*>?aƋĎ2*U#}•ĵAȴȵ*Ë°±>?ª–p
ö$$
 7 + 2 10 =
5 + 2 + 2 10
=
( 5)2 + ( 2)2 + 2 ×
5 ×
2
=
( 5 +
2)2
=
5 +
2}
Öëì‘
 5 +
2p
ý2 16-1_24-25 ĢßÃu·àáƪÿ·fïƑ`|2−3| +
4−2 3 = _    `
Ap2 3−3
Bp2 3−1
Cp3
Dp1
ëìD
OíÑĚěïðĜ012*¼T+,a"#© '*89P´Ë2ÜT012*<=°±>
?p
ö$$
|2−3| +
4−2 3
= |2−3| +
( 3)2−2 3 + 1
= 2−3 +
( 3−1)2
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
= 2−3 +
3−1
= 1a
Öô
Dp
ý2 16-2Ţ𝑎‘−13 +
5−3−5*J)NOa¦𝑎= _ƎƎ`
Ap
7
5
Bp
2
2
Cp 2−1
Dp 3−1
ëìC
Oíîïð-.)*/'p"#ȔĤËÀĸŗ2ýsaºÂ?>İ,Üĉa@Â/İ„J)NO*
'aºÂÄôõA°±ðŊa@×*ª‘ëìp
ö$$
îč:;a𝑎‘−13 +
5−3−5*J)NOaüÑ𝑎< 1aüіtņµ A D ôõp
Ţ𝑥=
5−3−5a
∵3−5 =
6−2 5
2
=
( 5−1)
2
2
a
∴3−5 =
( 5−1)
2
2
=
5−1
2 =
10−2
2
a
∴𝑥=
5−
10−2
2
=
2 5−10+ 2
2
a
∵5 ≈2.24a 10 ≈3.16a 2 ≈1.41a
∴𝑥=
2 5−10+ 2
2
= 2 × 2.24−3.16 + 1.41
2
= 4.48−3.16 + 1.41
2
2.73
2 = 1.365a
∵−13 +
5−3−5 = −13 + 𝑥= −11.635a
∵−11.635*M)NOz−12a
∴J)NO‘−11.635−(−12) = 0.365a
ôõ B z
2
2 ≈0.707aôõ C z 2−1 ≈0.41a
Ȏ7ôõ C @×ëìp
Öô
Cp
ý2 16-3_24-25 ŔßÛ·śľȶÿ·äå`>? 23−6 10 + 4 3−2 2*Üĉz_   `
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
Ap3 +
2
Bp3−2 2
Cp3 + 2 2
Dp3−2
ëìD
Oíîïð012*+,ač012*<=?ÜT"#ƋĎ2İ,ª–$p
ö$$
ŗ2 =
23−6 10 + 4 ( 2−1)
2
=
23−6 10 + 4 2−4
=
23−6 4 2 + 6
 =
23−6 (2 +
2)
2
=
23−12−6 2
=
11−6 2
=
(3−2)2
= 3−2a
Öô
Dp
【题型17  利用分母有理化求值】
£ 17_24-25 ŔßÛ·ŁłŽš·äA`ȷǫuÊİ,Ÿ%
1
2+1 =
1×( 2−1)
( 2+1)( 2−1) =
2−1}
1
3+ 2 =
1×( 3−2)
( 3+ 2)( 3−2) =
3−2}
1
5+2 =
1×( 5−2)
( 5+2)( 5−2) =
5−2p
Ƒ
(1)
1
7+ 6*'‘ p
(2)
1
1+ 2 +
1
2+ 3 +
1
3+ 4 +⋯+
1
2022+ 2023 +
1
2023+ 2024*'p
(3)¤𝑎=
1
5−2a𝑎3−4𝑎2 +𝑎+ 4*'p
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
ëì(1) 7−6
(2)2 506−1
(3)2 5 +8
OíÑïðĜOW7.> 012*>?'aȸȹOW7.>*Ëz$î*Ćp
_1`ŗ2čȷǫȺǓA*Ëýsª–ġÆÜĉ}
_2`ŗ2óõýsÂaȻȼTŖª–ġÆÜĉ}
_3`À>? aaºÂĶķĽ2Əİ,ª–p
ö$_1`
1
7+ 6 =
7−6
( 7+ 6)( 7−6) =
7−6}
_2`ŗ2 =
2−1 +
3−2 +
4−3 +⋯+
2023−2022 +
2024−2023
=
2024−1
= 2 506−1}
_3`∵𝑎=
1
5−2 =
5+2
( 5−2)( 5+2) =
5 +2a
∴𝑎−2 =
5a
∴(𝑎−2)2 = 5a
∴𝑎3−4𝑎2 +𝑎+ 4
= 𝑎(𝑎2−4𝑎) + 𝑎+ 4
= 𝑎(𝑎2−4𝑎+ 4)−3𝑎+ 4
= 5𝑎−3𝑎+ 4
= 2𝑎+ 4
= 2( 5 + 2) + 4
= 2 5 +8p
ý2 17-1ԟ»Ù+,aĸOWA*012>‘7.2*Ÿ%a‘OW7.>p
{
1
2 =
2
2 a
1
3−2 =
1×( 3+ 2)
( 3−2)_ 3+ 2`
=
3 +
2ȁ+,š‘OW7.>p
(1)ĸuæ012OW.>
2−3
12 = ___________a
𝑥−𝑦
𝑥−𝑦= ___________
(2)Ƚ Ⱦ…ȿ>?
𝑥−𝑦
𝑥+ 𝑦Ûaƒ„…wm¿*$ëŸ%p
Ƚ
𝑥−𝑦
𝑥+ 𝑦=
(𝑥−𝑦)
𝑥−𝑦
𝑥+ 𝑦
𝑥−𝑦=
(𝑥−𝑦)
𝑥−𝑦
𝑥−𝑦
=
𝑥−𝑦}
Ⱦ
𝑥−𝑦
𝑥+ 𝑦=
( 𝑥)
2−( 𝑦)2
𝑥+ 𝑦
=
𝑥+ 𝑦
𝑥−𝑦
𝑥+ 𝑦
=
𝑥−𝑦
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
ũĔɀɁȷǫȽ Ⱦ…ȿ*$Ÿ%a Ƚ Ⱦ…ȿ*$ëđ„ɂƁ_ƎƎ`
ApȽ Ⱦè    BpȽèȾ     CpȽ ȾĎ     DpȽ ȾĎè
(3)Ęę7.) a b ÷ø
𝑎
2+1 +
𝑏
2 = −1 + 2 2a a b *'p
ëì(1)
2 3−3
6
a 𝑥+
𝑦
(2)C
(3)𝑎= 1a𝑏= 2
OíîĚěïðĜOW7.>a$*ĆzŊ”7.>ü2p
_1`č´t7.>ü2Cčň2ü2O$>?İ,ª–}
_2`č_1`AË°±ƁƂª–}
_3`čËba°±OW7.>İ,ġ„ 𝑎+
1
2 𝑏
2−𝑎= −1 + 2 2ač𝑎,𝑏‘7.)a°·ª–ġ
𝑎,𝑏*'aª–$p
ö$_1`$
2−3
12 =
2 3−3
2 3× 3 =
2 3−3
6
𝑥−𝑦
𝑥−𝑦
=
(𝑥−𝑦)
𝑥+
𝑦
𝑥−𝑦
𝑥+
𝑦
= (𝑥−𝑦)
𝑥+
𝑦
𝑥−𝑦
=
𝑥+
𝑦
Öëì‘
2 3−3
6
a 𝑥+
𝑦}
_2`č_1`A*Ë°±İ,–ęaȽ Ⱦš 
Öô
Cp
_3`$
 ∵
𝑎
2+1 +
𝑏
2 = −1 + 2 2
∴( 2−1)𝑎+
2
2 𝑏= −1 + 2 2
∴
2𝑎−𝑎+
2
2 𝑏= −1 + 2 2
∴𝑎+
1
2 𝑏
2−𝑎= −1 + 2 2
∵𝑎a𝑏 z7.)
∴𝑎+ 1
2 𝑏= 2a−𝑎= −1
∴𝑎= 1
∴1 + 1
2 𝑏= 2
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∴𝑏= 2p
ý2 17-2_24-25 ŔßÛ·ĬɃ]·äA`ȷǫȺǓ
«°±012>?ÛaȚī7ÛȈɄ›{
3
5a
2
3a
2
3+1Čǥb’*2Əa6(ȚīӖtĸ6°bɅ>
?
3
5 =
3× 5
5× 5 =
3
5 5}
2
3 =
2×3
3×3 =
6
3 }
2
3+1 =
2×( 3−1)
( 3+1)( 3−1) =
2( 3−1)
( 3)
2−12 =
3−1pt›Čw>?*ɅɆd
eOW7.>p
$ƤĦ
(1)ɇĵ›Ê*$Ÿ%a>?
1
7−6 = __________p
(2)İ,
1
2+1 +
1
3+ 2 +
1
4+ 3 + ⋯+
1
2025+ 2024 × ( 2025 + 1)p
(3)Ęę𝑎=
1
3+ 2a𝑏=
1
3−2a𝑎2 + 𝑏2*'p
ëì(1) 7 +
6
(2)2024
(3)10
OíîïðOW7.)a012*¼T+,a>?'
_1`čOW7.>*Ë°±$ª–}
_2`À°±OW7.>aÁ°±İ,ª–}
_3`À°±OW7.>„𝑎,𝑏*'a°·„𝑎+ 𝑏,𝑎𝑏*'a"#ƋĎ2ýs'ª–p
ö$_1`$
1
7−6 =
7+ 6
( 7−6)( 7+ 6) =
7+ 6
7−6 =
7 +
6}
Öëì‘
 7 +
6}
_2`ŗ2 = (−1 +
2−2 +
3 + ⋯−2024 +
2025) × ( 2025 + 1)
= ( 2025−1) × ( 2025 + 1)
= 2025−1 = 2024}
_3`∵𝑎=
1
3+ 2 =
3−2
( 3+ 2)( 3−2) =
3−2a
𝑏=
1
3−2 =
3+ 2
( 3−2)( 3+ 2) =
3 +
2a
∴𝑎+ 𝑏= 2 3a𝑎𝑏= 1a
∴𝑎2 + 𝑏2 = (𝑎+ 𝑏)2−2𝑎𝑏= (2 3)
2
−2 = 10p
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ý2 17-3_24-25 ŔßÛ·GɈ·äA`£{
2+1
2−1 =
( 2+1)( 2+1)
( 2−1)( 2+1) = 2 2 +3pɉČǥԟOƏ OW
¿´bc2ƏŀOWA*	>ǝCŚŀ	A*OW>ǝadeOW7.>p7uæÜÝ
①¤ a z 3*J)NOa¦
3
𝑎*'‘3 3 +6}
②KLJ
 2024−2023 <
2023−2022}
③ýs
1
3+ 2+1 =
2+ 2−6
4
}
④İ,
2
3+ 3 +
2
5 3+3 5 +
2
7 5+5 7 +⋯+
2
99 97+97 99 = 1−
11
33 }
⑤Ęę𝑎=
𝑚+1−𝑚
𝑚+1+ 𝑚a𝑏=
𝑚+1+ 𝑚
𝑚+1−𝑚a¸𝑎2 +1926𝑎𝑏+ 𝑏2 = 2024a¦Ľ7–‹*M) m *P‘−1p
t›ÜݗÕ*z_    `
Ap①③④⑤
Bp②③④
Cp②③⑤
Dp②③④⑤
ëìB
Oíîïð"#O2*Ɋî<= ň2°±OW7.>a$Ƥ012*>? KLJP
+,*ĦaĠĄćĈęļœz$*Ćp
①/İ-.)*M)NOa„J)NO𝑎=
2−1a°·OW7.>°±>?}
②ԟOW7.>aKL…c012*§)Jaª–$ë}
③ÀOƏOW¿Û´t( 3−2−1)a³ŪOW*2c)ÂÁ17.>OWª–}
④ԟOW7.>ŊÆA-.2P*+,XYaÄ·>?„'}
⑤𝑎Q b –t"#OW7.>>?a–ġ„𝑎𝑏= 1aºÂȲȳ%ɋœaġ m *'p
ö$$
①∵a z 3*J)NOa
∴𝑎=
3−1a
∴
3
𝑎=
3
3−1 =
3( 3+1)
( 3−1)( 3+1) =
3( 3+1)
2
=
3 3+3
2
aÖ①èé}
②∵
1
2024−2023 =
( 2024−2023)( 2024+ 2023)
( 2024−2023)
=
2024 +
2023a
1
2023−2022 =
( 2023−2022)( 2023+ 2022)
( 2023−2022)
=
2023 +
2022a
Ɩ∵2024 +
2023 >
2023 +
2022a
∴
1
2024−2023 >
1
2023−2022a
∴2024−2023 <
2023−2022aÖ②—Õ}
③
1
3+ 2+1
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
=
( 3−2−1)
( 3 +
2 + 1)( 3−2−1)
=
3−2−1
3−( 2 + 1)2
=
3−2−1
−2 2
=
2( 3−2−1)
−2 2 ×
2
=
2+ 2−6
4
aÖ③—Õ}
④∵
2
(2𝑘+3) 2𝑘+1+(2𝑘+1) 2𝑘+3
=
2[(2𝑘+ 3) 2𝑘+ 1−(2𝑘+ 1) 2𝑘+ 3]
[(2𝑘+ 3) 2𝑘+ 1 + (2𝑘+ 1) 2𝑘+ 3][(2𝑘+ 3) 2𝑘+ 1−(2𝑘+ 1) 2𝑘+ 3]
= 2[(2𝑘+ 3) 2𝑘+ 1−(2𝑘+ 1) 2𝑘+ 3]
(2𝑘+ 3)2(2𝑘+ 1)−(2𝑘+ 1)2(2𝑘+ 3)
= 2[(2𝑘+ 3) 2𝑘+ 1−(2𝑘+ 1) 2𝑘+ 3]
(2𝑘+ 3)(2𝑘+ 1)[(2𝑘+ 3)−(2𝑘+ 1)]
= (2𝑘+ 3) 2𝑘+ 1−(2𝑘+ 1) 2𝑘+ 3
(2𝑘+ 3)(2𝑘+ 1)
=
2𝑘+ 1
2𝑘+ 1 −
2𝑘+ 3
2𝑘+ 3
∴
2
3+ 3 +
2
5 3+3 5 +
2
7 5+5 7 + ⋅⋅⋅+
2
99 97+97 99
=
2(3−3)
(3+ 3)(3−3) +
2(5 3−3 5)
(5 3+3 5)(5 3−3 5) +
2(7 5+5 7)
(7 5+5 7)(7 5−5 7) +⋯+
2(99 97−97 99)
(99 97+97 99)(99 97−97 99)
=
1−
3
3
+
3
3 −
5
5
+
5
5 −
7
7
+ ⋯+
97
97 −
99
99
= 1−
3
3 +
3
3 −
5
5 +
5
5 −
7
7 + ⋯+
97
97 −
99
99
= 1−
11
33 aÖ④—Õ}
⑤∵𝑎=
𝑚+1−𝑚
𝑚+1+ 𝑚=
( 𝑚+1−𝑚)( 𝑚+1−𝑚)
( 𝑚+1+ 𝑚)( 𝑚+1−𝑚) = ( 𝑚+ 1−𝑚)
2
a
∴𝑎> 0a
∵𝑏=
1
𝑎a
∴𝑎𝑏= 1a𝑏=
𝑚+1+ 𝑚
𝑚+1−𝑚= ( 𝑚+ 1 +
𝑚)
2
a
∴𝑏> 0a
∴𝑎+ 𝑏> 0a
∵𝑎2 +1926𝑎𝑏+ 𝑏2 = 2024a
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
∴𝑎2 +1926 + 𝑏2 = 2024a
𝑎2 + 𝑏2 = 98a
𝑎2 + 𝑏2 +2𝑎𝑏= 98 + 2𝑎𝑏a
(𝑎+ 𝑏)2 = 98 + 2a
(𝑎+ 𝑏)2 = 100a
∵𝑎+ 𝑏> 0a
∴𝑎+ 𝑏= 10a
ª( 𝑚+ 1−𝑚)
2
+ ( 𝑚+ 1 +
𝑚)
2
= 10a
4𝑚+ 2 = 10a
$ġ𝑚= 2aÖ⑤èép
S›ĽɌ
②③④—Õa
Öô
Bp
【题型18  二次根式的规律探究】
£ 18>?
1
8+ 11 +
1
11+ 14 +
1
14+ 17 +
1
17+ 20 +
1
20+ 23 +
1
23+ 26 +
1
26+ 29 +
1
29+ 32Üĉz_   `
Ap1
Bp
2 2
3
Cp2 2
Dp4 2
ëìB
OíîïðĜOW7.>5012*²³+,aĠĄćĈ+,˦z$*ĆpÀĸ™cO2
°±OW7.>aÁİ,²³ª–ġ„ëìp
ö$$
1
8+ 11 =
11−8
( 11+ 8)( 11−8) =
11−8
3
a
¿.–ġ
1
11+ 14 =
14−11
3
⋅⋅⋅
1
29+ 32 =
32−29
3
a
1
8 +
11
+
1
11 +
14
+
1
14 +
17
+
1
17 +
20
+
1
20 +
23
+
1
23 +
26
+
1
26 +
29
+
1
29 +
32
=
11−8
3
+
14−11
3
+
17−14
3
+
20−17
3
+
23−20
3
 +
26−23
3
+
29−26
3
+
32−29
3
=
32−8
3
= 4 2−2 2
3
= 2 2
3
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Öô Bp
ý2 18-1_24-25 ĢßÃu·àáɍǺ·äå`Ȳȳuæ|2
①3 + 2 2 = ( 2 + 1)
2
}②5 + 2 6 =
( 3 +
2)
2
}③7 + 2 12 = (2 +
3)
2
}…}ũčt›XYaƒ„Ň 9 c|2      
p
ëì19 + 2 90 = ( 10 + 3)
2
OíîïðĜ)ʼn’XYZɎa012*<=ačĘę|2ĕĖbɏXYz$ĆpȲȳ|
2–ġŇ𝑛c|2‘(2𝑛+ 1) +2 𝑛(𝑛+ 1) = ( 𝑛+ 1 +
𝑛)
2
aª–$p
ö$$
&8–ęaŇ 1 c|23 + 2 2 = ( 2 + 1)
2
aª(2 × 1 + 1) +2 1 × 2 = ( 2 +
1)
2
}
Ň 2 c|25 + 2 6 = ( 3 +
2)
2
aª(2 × 2 + 1) +2 2 × 3 = ( 3 +
2)
2
}
Ň 3 c|27 + 2 12 = (2 +
3)
2
aª(2 × 3 + 1) +2 3 × 4 = ( 4 +
3)
2
}
……
ȲȳĕĖaŇ𝑛c|2‘(2𝑛+ 1) +2 𝑛(𝑛+ 1) = ( 𝑛+ 1 +
𝑛)
2
a
¦Ň 9 c|2‘19 + 2 90 = ( 10 + 3)
2
a
Öëì‘
19 + 2 90 = ( 10 + 3)
2
p
ý2 18-2ԟ“&ɋɐÆbɏ”*ËZ[uÊ012*+,XY
ɋ£ 1
 1 +
1
3 =
3+1
3 =
4 ×
1
3 = 2
1
3}ɋ£ 2
 2 +
1
4 =
8+1
4 =
9 ×
1
4 = 3
1
4}ɋ£ 3
 3 +
1
5 =
15+1
5
=
16 ×
1
5 = 4
1
5……U#ĕĖ*+
,XY 2024 +
1
2026 ×
4052*'_      `
Ap2024
Bp2025 2
Cp2023
Dp2023 2
ëìB
OíîïðŊXYačAɋ£ġÆXY 𝑛+
1
𝑛+2 =
𝑛(𝑛+2)+1
𝑛+2
=
(𝑛+ 1)2 ×
1
𝑛+2 = (𝑛+ 1)
1
𝑛+2a
Ķ'$ª–ġÆëìaȲȳĘęɋ£ɋɑa”ÕŊÆXYz$ƤĦ*Ćp
ö$$
ɋ£ 1
 1 +
1
3 =
3+1
3 =
4 ×
1
3 = 2
1
3}
ɋ£ 2
 2 +
1
4 =
8+1
4 =
9 ×
1
4 = 3
1
4}
ɋ£ 3
 3 +
1
5 =
15+1
5
=
16 ×
1
5 = 4
1
5}
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
……
∴t›XY‘
 𝑛+
1
𝑛+2 =
𝑛(𝑛+2)+1
𝑛+2
=
(𝑛+ 1)2 ×
1
𝑛+2 = (𝑛+ 1)
1
𝑛+2a
∴Ù𝑛= 2024Ûa 2024 +
1
2026 ×
4052 = 2025
1
2026 ×
4052 = 2025 2a
Öô
Bp
ý2 18-3_24-25 ŔßÃu·àáƌÿ·äA`ÀȲȳuæ|2aÁŎëĦ
①1 +
1
12 +
1
22 = 1 +
1
1−
1
2}
②1 +
1
22 +
1
32 = 1 +
1
2−
1
3}
③1 +
1
32 +
1
42 = 1 +
1
3−
1
4}
…
ũĔ"#ĕĖ*XYaİ,
1 +
1
12 +
1
22 +
1 +
1
22 +
1
32 +
1 +
1
32 +
1
42 +⋯+
1 +
1
20222 +
1
20232−2024 =      
p
ëì−
2024
2023/−1
1
2023
Oí①1 +
1
12 +
1
22 = 1 +
1
1−
1
2}②1 +
1
22 +
1
32 = 1 +
1
2−
1
3}③1 +
1
32 +
1
42 = 1 +
1
3−
1
4aġÆ
1 +
1
20222 +
1
20232 = 1 +
1
2022−
1
2023aæ2İ,ª–p
îïðĜ012AXYZɎa()*İ,aĠĄćĈXYZɎz$*Ćp
ö$$
①1 +
1
12 +
1
22 = 1 +
1
1−
1
2}
②1 +
1
22 +
1
32 = 1 +
1
2−
1
3}
③1 +
1
32 +
1
42 = 1 +
1
3−
1
4a
Ö 1 +
1
20222 +
1
20232 = 1 +
1
2022−
1
2023a
Ö 1 +
1
12 +
1
22 +
1 +
1
22 +
1
32 +
1 +
1
32 +
1
42 +⋯+
1 +
1
20222 +
1
20232−2024
=
1 + 1 + 1 + … + 1
⏟
2022
+ 1−1
2 + 1
2 −1
3 + 1
3 −1
4 + … +
1
2022 −
1
2023 −2024
=
1 + 1 + 1 + … + 1
⏟
2022
+ 1−1
2 + 1
2 −1
3 + 1
3 −1
4 + … +
1
2022 −
1
2023 −2024
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
= 2023−
1
2023 −2024
= −
2024
2023a
Öëì‘
−
2024
2023p
【题型19  二次根式的新定义】
£ 19_24-25 ŔßÃu·śľɒɓ·Ă㥹`]9
Țīĸ( 𝑎+
𝑏)Q( 𝑎−𝑏)‘b “ ǎ2”pü
‘
( 𝑎+
𝑏)( 𝑎−𝑏) = ( 𝑎)
2
−( 𝑏)
2
= 𝑎−𝑏a–t7ɔ*ǝɕ	aĽt7bĦ–tԟɖǖ“ ǎ2”
$Ƥp£{
Ęę 18−𝑥−11−𝑥= 1a 18−𝑥+
11−𝑥*'a–tČǥ$ë
ü‘( 18−𝑥−11−𝑥) × ( 18−𝑥+
11−𝑥) = ( 18−𝑥)
2
−( 11−𝑥)
2
= 18−𝑥−11 + 𝑥 = 7aƖü‘ 18−𝑥−
11−𝑥= 1aĽt 18−𝑥+
11−𝑥= 7p
_1`Ęę
 20−𝑥+
4−𝑥= 8a¦𝑥*'z           
}
_2`İ,
1
3 1+ 3 +
1
5 3+3 5 +
1
7 5+5 7 +⋯+
1
2023 2021+2021 2023 =           
ëì     
−5     
1
2−
2023
4046
OíÑïðĜ012*<= 012*¼T+, 012789*:; ň2t5O
W7.>aĠĄćĈ012*+,˦PƯưýsz$*Ćp
_1`ɇĵ£aæ„%Ša$%Šª–ġÆëì}
_2`"#ň2a ŗ2°±ýsÂaª–ġÆëìp
ö$$
∵( 20−𝑥+
4−𝑥)( 20−𝑥−4−𝑥) = ( 20−𝑥)
2
−( 4−𝑥)
2
= 20−𝑥−4 + 𝑥= 16a
¸ 20−𝑥+
4−𝑥= 8a
∴20−𝑥−4−𝑥= 2}
∵
20−𝑥+
4−𝑥= 8
20−𝑥−4−𝑥= 2
 
∴2 4−𝑥= 6a
>?…ϿÛġ
4−𝑥= 9a
∴𝑥= −5a
ůƟɗ
𝑥= −5zŗ%*$}
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
Öëì‘
−5p
_2`
1
3 1+ 3 +
1
5 3+3 5 +
1
7 5+5 7 + ⋅⋅⋅+
1
2023 2021+2021 2023
= 3−3
6
+ 5 3−3 5
30
+ 7 5−5 7
70
+⋅⋅⋅+
2023 2021−2021 2023
20232 × 2021−20212 × 2023
= 1
2 −
3
6 +
3
6 −
5
10 +
5
10 −
7
14 +⋅⋅⋅+
2021
4042 −
2023
4046
=
1
2−
2023
4046 p
Öëì‘
1
2−
2023
4046 p
ý2 19-1_24-25 ŔßÃu·àáƌÿ·äA` Ík8my|*…c()𝑎,𝑏a]9+,※{u
Ù𝑎< 𝑏
Ûa𝑎※𝑏= 2 𝑎+
𝑏aÙ𝑎≥𝑏Ûa𝑎※𝑏= 2 𝑎−𝑏a£{5※2 = 2 5−2a•›ɌX]aİ, 3※2 −
8※12 *Üĉ‘_   `
Ap4 3−5 2
Bp−5 2
Cp4 3−3 2
Dp3 2
ëìB
Oíîïð*z()*+,ačĽȖ*2Ə„3※2P8※12*'aÁč012*²³İ,
Ë°±İ,ª–p
ö$$
&8ġa
3※2 = 2 3−2a
8※12 = 2 8 +
12 = 4 2 +2 3a
3※2 −8※12
= 2 3−2−(4 2 + 2 3)
= 2 3−2−4 2−2 3
= −5 2a
Öô
Bp
ý2 19-2_24-25 ŔßÃu·ŁłŃń·Ă㥹`]9
ü‘( 𝑎+
𝑏)( 𝑎−𝑏) = ( 𝑎)
2
−( 𝑏)
2
= 𝑎−𝑏a–t7ɔ*ǝɕ	aȚī( 𝑎+
𝑏)Q( 𝑎−𝑏)‘b “ ǎ2”p¤ 18−𝑥−11−𝑥= 1a¦
18−𝑥+
11−𝑥=     p
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
ëì7
OíîĚěïð012*+,aĠĄćĈ012*´Ë+,5AĽȖ+,z$*Ćpɘ
ę 18−𝑥+
11−𝑥Q 18−𝑥−11−𝑥zb “ ǎ2”a–č 18−𝑥+
11−𝑥= ( 18−𝑥+
11−𝑥)
( 18−𝑥−11−𝑥)>?İ,ª–p
ö$$
čȺǓ–ęa 18−𝑥+
11−𝑥Q 18−𝑥−11−𝑥zb “ ǎ2”a
∵18−𝑥−11−𝑥= 1a
∴18−𝑥+
11−𝑥= ( 18−𝑥+
11−𝑥)( 18−𝑥−11−𝑥)
= ( 18−𝑥)2−( 11−𝑥)2
= 18−𝑥−(11−𝑥)
= 7
Öëì‘
7p
ý2 19-3 Ík8() mana¤]9\+,𝑚⊗𝑛=
𝑚−𝑛(𝑚≥𝑛),
𝑚+
𝑛(𝑚< 𝑛)
 
aȖ„ŋcçË
①18 ⊗2 = 2 2}
②
1
1⊗2 +
1
2⊗3 +
1
3⊗4 + ⋅⋅⋅+
1
99⊗100 = 100 ⊗1}
③(𝑎⊗𝑏) ⋅(𝑏⊗𝑎) = |𝑎−𝑏|p
t›çËA—Õ*c)z_    `
Ap0 c
Bp1 c
Cp2 c
Dp3 c
ëìD
Oí"#\]9°±İ,òbƁƂª–p
ö$$
∵18 > 2a
∴18 ⊗2 =
18−2 = 3 2−2 = 2 2a
Ľt①—Õ}
1
1 ⊗2 +
1
2 ⊗3 +
1
3 ⊗4 +⋅⋅⋅+
1
99 ⊗100
=
1
1 +
2
+
1
2 +
3
+
1
3 +
4
+ ⋯+
1
99 +
100
=
2−1 +
3−2 + ⋯+
100−99
=
100−1
= 100 ⊗1
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
Ľt②—Õ}
Ù𝑎≥𝑏Ûa(𝑎⊗𝑏) ⋅(𝑏⊗𝑎) = ( 𝑎−𝑏)( 𝑏+
𝑎) = 𝑎−𝑏= |𝑎−𝑏|a
Ù𝑎< 𝑏Ûa(𝑎⊗𝑏) ⋅(𝑏⊗𝑎) = ( 𝑎+
𝑏)( 𝑏−𝑎) = 𝑏−𝑎= |𝑎−𝑏|a
Ľt③—Õ}
֗Õ*‘①②③a7 3 ca
Öô Dp
œĥîïð\]9a012*¼T+,aćĈ\]9*+,˦z$*Ćp
【题型20  二次根式的实际应用】
£ 20{ŅaĸbǑəǮɚy–t­ŽbcĨ‘ 
8𝜋aȣ‘ 
2𝜋*ɛsa¤ĸČǑəƢfG\Ǯ
ɚy­Žbc~sa¦Ũ~*ÊƓz _      `
Ap12π
Bp18π
Cp24π
Dp36π
ëìB
OíîïðĜ012*¼T+,aɛs*<=a~*ÊƓ2ač8ġ„Ĩa°·ġ~
*ljNJač~*ÊƓ2aª–$p
ö$$
ČǑə*Ĩ‘2( 8π +
2π) = 6 2π
∴ĸČǑəƢfG\Ǯɚy­Žbc~sa¦ljNJ‘
6 2π
2π = 3 2
∴ÊƓ‘π(3 2)
2
= 18π
Öô
Bp
ý2 20-1{Ņa«—sɜŹAǢƻ…cJ—saĘę…cJ—s*ÊƓOi‘𝑆1 = 18a𝑆2
= 12aGɝNOzbc—sa6ÊƓ‘ 2a¦ėɞNO*ÊƓ‘      
p
ëì8 6−8/−8 + 8 6
OíîïðĜ012*U#aÀ,„ŋcJ—s*ÏĨaÁġÆ—s*ÏĨaԟÊƓ*
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
İ,ġÜÝp
ö$$
 ∵ŋcJ—s*ÊƓOi‘ 18 12 2a
∴ŋcJ—s*ÏĨOi‘ 18 = 3 2 12 = 2 3 2a
&Ņę
—s*ÏĨ‘
3 2 +2 3−2 = 2 2 +2 3a
∴𝑆空白= (2 2 + 2 3)2−(18 + 12−2)
= 8 + 12 + 8 6−(18 + 12−2)
= 8 6−8p
Öëì‘
8 6−8p
ý2 20-2_24-25 ŔßÛ·àĺɟA·äA`ĕɠNjÔɡÖÂaNjɢɣȳÔqčɤɥÂɥōɦŸ*ǀ
ǁ/İɥɧ±ɨ*ɩǣaĽ#*ůɗ2z𝑣= 16 𝑑𝑓a6A𝑣Œɥɩ_ǼŠ
km h`a𝑑ŒɤɥÂɥ
ōɦŸ*ǀǁ_ǼŠ
m`a𝑓Œɪɫü)p«Ǹ1NjÔɡÖɬðAaɭġ𝑑= 20ma𝑓= 1.2a¦ɮɡɯ
ɥ*ɥɩɰz‰Ūū_ 6 ≈2.45aÜĉɱÕÆ1km/h`
ëìɮɡɯɥ*ɥɩɰz78km/h
OíîĚěïðĜ012*U#a>?012|ęļœaĸ𝑑= 20a𝑓= 1.2Ķķ𝑣= 16 𝑑𝑓ª
–„ɮɡɯɥ*ɥɩɰz‰ŪaĠĄćĈ+#012*<=>?'z$ƤÑ*Ćp
ö$$
∵𝑑= 20a𝑓= 1.2Ķķ𝑣= 16 𝑑𝑓a
∴𝑣= 16 ×
20 × 1.2 = 16 × 2 6 = 32 6a
∵6 ≈2.45a
∴𝑣≈32 × 2.45 ≈78(km h)a
ë
ɮɡɯɥ*ɥɩɰz78km/hp
ý2 20-3_24-25 ŔßÛ·ľĺāɲ·äå`ǘėɳɴzbwmɵŮ*ɶɷ±‘ačɸ[aɴɹÄǁȤ
ʑ h Ǫ*ǘdžŶ&ɺuaɺÆȤÊ*ۇ‘ taůŸ(ɗaĕĖ𝑡=
ℎ
5(mïɻɼƇ*ɽɾ)p
(1)ɴƗÄ40m*ǘėɺÆȤÊ*ۇ}
(2)ĘęÄǘėɿɺ*ɴƗĽÚ‹ʀ(ǼŠ
J) = 10 × ɴƗ=ʀ(kg) × ǘǣ(m)abʁ=ʀ‘0.2kg*ʂʃůŸ
3sɺ«Ȥ›aČʁʂʃ«uɺŸ%AĽÚ‹ʀ7‰ū
(3)«_2`*ÜĉAaĔ‹ġÆʄċʅū(
ʆʇ-ʈʉȿƗȎñě65J*‹ʀ)
ëì(1)2 2s
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
(2)90J
(3)ʊʋǘėɳɴ
OíîïðĜĶ)2*'a012*U#a—Õ.$8z$*Ćp
_1`ŀℎ= 40mĶķ𝑡=
ℎ
5İ,ª–}
_2`𝑡= 3sĶķ𝑡=
ℎ
5ġ ha°·İ,ª–}
_3`&Í90J>65JaÖȈ ȿƗǖŽʇʌa¦UŨʋűǘėɳɴ
ö$_1`$
ŀℎ= 40mĶķ𝑡=
ℎ
5ġ
𝑡=
40
5 = 2 2sa
ë
ɴƗÄ40m*ǘėɺÆȤÊ*ۇ‘2 2s}
_2`$
𝑡= 3sĶķ𝑡=
ℎ
5ġ
ℎ
5 = 3a
$ġ
ℎ= 45ma
¦Äǘėɿɺ*ɴƗĽÚ‹ʀ‘10 × 0.2 × 45 = 90Ja
ë
Čʁʂʃ«uɺŸ%AĽÚ‹ʀ790J}
_3`$
∵90J>65Ja
∴ ȿɖŽʇʌa
Öʊʋǘėɳɴp
教辅资源，关注公众号★全科AA+