第二章实数的初步认识(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(苏科)

                                                                           
                                                                       
第二章  实数的初步认识（举一反三讲义）全章题型归纳
 2024
 ...................................................................................................................................................................5
 1    ........................................................................................................................5
 2  !" #$ ............................................................................................................6
 3  % & ........................................................................................................................9
 4  '()*+ ..........................................................................................................................................11
 5  ,-().+ ......................................................................................................................................13
/0 .................................................................................................................................................................15
 6  +1)23456 ..................................................................................................................15
 7  '()789: ..................................................................................................................................17
 8  ;(<=>8(<= ......................................................................................................................19
 9  '(?(@AB*" ..........................................................................................................................21
 10 '(*+)C" ..................................................................................................................................24
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  
!


"#


#$
%&'
()*+,-#
./0
 a ( 𝑎
 a ./( ±
𝑎
12
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.KF12M=>
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
]^
;_
/`
𝒂²
𝒂𝟐=|𝒂|= 𝒂D𝒂≥𝟎E
−𝒂D𝒂< 𝟎E
62=|6|=6
(−6)2=|−6|=6
𝒂𝟐
( 𝐚)𝟐 = 𝐚D𝐚≥𝟎E
( 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=aF x
 a ab
 !



 x 
 a, x²=a, x 
 a  
!

c
 a #d"#
a

aB
,
aB,

#$
%&
'()*+,-#
./0
3 𝑎
±
𝑎
12
𝑎3456
ef
𝑎≥0
)gh
E 0
0 B 0
0 aB 0
知识点5  开立方
1.GH( a ) )*+IJK FLM a IJNK(.
2.K>K )OP
K
K 
*+Q	
±
3
NK(
:,
RS(
                                                                           
                                                                       
H(
0 "#
+

#$+,-#
efaG"#


知识点6  无理数
,TUVW8(IJ,-(X
,-()YZ[\]^_`a
D1EKKUb)()cCd,-(Fe 2F
3 5fg
D2EhijπkGlm]π)(,e 2πF
π
2fg
D3E^,TUVW8([\no)(,e 0.101 001 000 1…DpH 1 qrstuGH 0)fXS,-()8
(<=v(,Tg,-()8(<=UVWg,-(Uwxyz=()[\X
知识点7  实数的概念及分类
1. {|
]-(>,-(}~'(X
2. =€
'(]pa=€‚
D1Eƒ„…=€
'(†=]-(>,-(X
D2Eƒ‡56=€
'(†=‡'(05'(X
知识点8  实数与数轴的关系
‡;(
‡=(
5;(
5=(
'(
‡'(
5'(
‡]-(
‡,-(
5]-(
5,-(
0
'(
]-(
0
,-(
‡]-(
5]-(
‡,-(
5,-(
]T8(‰VW8(
,TUVW8(
‡;(
‡=(
5;(
5=(
                                                                           
                                                                       
ŠGH'(‹†^"(@Œ)GHŽxygŽF(@Œ)ŠGH‹xyGH'(X'(?(@Œ
)GG‘C.
知识点9  实数范围内的有关概念
’~
6“
”•
c(
– a ? b —c(F˜𝑎+ 𝑏= 0
3)c(d−3
™(
– a ? b —™(F˜𝑎𝑏= 1
2 )™(d
1
2
Rš'()›‘&‹d45(Fœ|𝑎| ≥0
|𝑎|
𝑎(𝑎> 0)
0(𝑎= 0)
−𝑎(𝑎< 0)
›‘&
—c()pH()›‘&cfFœ|𝑎| = |−𝑎|
| 2| = |−2| =
2
知识点10  实数的运算
'(žŸ FU¡†^¢£¤¥¦§D§(U¨E¦*+F©ª†^¢£K *+^k
45'()K*+X
]-()*+6“>*+«'(žŸ ¬­®"F'(¯B*+)°±?]-(¯B*+)°±c²F
³¦KF´¦§Fµ¶¤¥X²·*+ƒ¸¹º»)°±¢£F]¼	)³+¼	½¾)X
知识点11  实数的大小比较
]-(789:)¿'(žŸ ¬­®"X
p'()78Àe_
‡'(‹7Á 0F5'(‹8Á 0F‡'(7ÁGÂ5'(gpH‡'(F›‘&
7)‡'(7gpH5'(F›‘&7)5'(8g(@Œxy)pH'(F»Ã)(Ä9¹Ã)(
7XÅÆF9:p'()78Ç]e_¿
D1EȐ9:p'())78F¢©É„'()78ÀXe9: 10? 3 )78F%Á( 10)
2 = 10
，32 = 9，10 > 9FÊ 10 > 3X
D2E".+)¿o,-()ËÌ&¶F´9:p()78X
D3EÍpHÎ	),-(9:78ÏF†C"e_ÐÑ
① 𝑎> 𝑏≥0⇔𝑎>
𝑏X
② 𝑎> 𝑏⇔
3 𝑎>
3 𝑏X
                                                                           
                                                                       
【培优篇】
【题型1  直接求平方根、立方根】
• 1D24-25 ÒÓ·_·ÔÕÖ×·ØÙE_ÚÛ¿MÜÝ)dD    E
AX 5d 5 )+1
BX0 )> ‹d 0
CX 9)d±3
DX−3d(−3)2)GH
ÞßC
=àáâã?+1){|FäåG=àæçè)‡É6X
é##
çè A
 5d 5 )+1F
+1d45(Fª 5êë( 5)
2
= 5FÊ A ‡Ég
çè B
0 )> ‹d 0F
> )„…MF0 )> ì 0FÊ B ‡Ég
çè C
 9)d ± 3F
9 = 3F© 3 )C ±
3F©4 ± 3F ± 3)d 9FíÁ 9)¯îÜÝFÊ C ÜÝg
çè D
−3d(−3)2)GHF
(−3)2 = 9F9 ) ± 3FïÅ−3d 9 )GHFÊ D ‡Ég
Êç CX
ð\ 1-1D24-25 ÒÓ·_·ñòóÕô·õö÷øE−
8
27) d   g 4)+1d     X
Þß     
−
2
3     
2
=àáâã F+1F#)ùdúû
eüGH‡(𝑥)fÁ𝑎FýþÿH‡(𝑥
IJ𝑎)+1geüGH(𝑦) fÁ𝑏FýþÿH(𝑦IJ𝑏) XĀÅ#ޜ†X
é##
−
8
27) d−
2
3g
∵4 = 2F2)+1d 2F
∴4)+1d 2X
ÊÞß
−
2
3g 2X
ð\ 1-2D24-25 ÒÓ·_·āĂ·ăØĄąED1EĆx
a
0.000001
0.001
1
1000
1000000
                                                                           
                                                                       
3 𝑎
D2EĀćĈĉ)Ċ«Ćċ
①
3 3 = 1.442F˜
3 3000 = ______F
3 0.003 = ______g
②Čč
3 343 = 7F˜
3 0.000343 = ______X
ÞßD1E0.01F0.1F1 F10 F100D2E①14.42F0.1442F ②0.07
=àáĎďâãĐ )6“FsĀNK(8(đ¹‰đ»Ēē 3 v‘C) )8(
đ¹‰đ»Ēē 1 v#œ†FĔ÷úûNK(8(?‘C) 8(ĒēĊ«d#)ùX
D1E!" )6“#œ†g
D2E①!" )6“#œ†g
②!" )6“#œ†X
é##
D1E
3 0.000001 = 0.01g
3 0.001 = 0.1g
3 1 = 1g
3 1000 = 10g
3 1000000 = 100g
ÊÞß
0.01F0.1F1 F10 F100g
D2E①
3 3000 =
3 1000 ×
3 3 = 10 × 1.442 = 14.42g
3 0.003 =
3 0.001 ×
3 3 = 0.1 × 1.442 = 0.1442g
ÊÞß
14.42F0.1442g
②
3 0.000343 =
3 0.000001 ×
3 343 = 0.01 × 7 = 0.07
ÊÞß
0.07X
ð\ 1-3– ab —c(Fcd —5™(F˜ 𝑎2−𝑏2 +
3 𝑐𝑑=       
.
Þß―1
é#ĀSĕ
a+b=0Fcd=-1F
˜ 𝑎2−𝑏2 +
3 𝑐𝑑= (𝑎+ 𝑏)(𝑎−𝑏) +
3 −1 =
0−1=-1.
ÊÞßd:-1.
【题型2  利用求平方根、立方根解方程】
• 2D25-26 ĖÓ·Œ·āĂ·ė¶ĄąE‘Á'(𝑎,𝑏F„…Ga*+“ ⊕”
𝑎⊕𝑏= 𝑎2−2𝑏XÁ𝑥)
                                                                           
                                                                       
$(𝑥+ 1) ⊕ 8 = 0)#            
X
Þß𝑥1 = 3,𝑥2 = −5
=àĀS„…)Ę*+FÚoÁ𝑥)GęĚt$œ†F#qœ†ĕºÞß.
é#%S†ĕ(𝑥+ 1)2−16 = 0F
∴(𝑥+ 1)2 = 16,
∴𝑥+ 1 = 4‰𝑥+ 1 = −4,
∴𝑥1 = 3,𝑥2 = −5X
ÊÞß
𝑥1 = 3,𝑥2 = −5.
ěáâã)dGęĚt$#)ĜF'()*+>-#„…)Ę*+d#)ù.
ð\ 2-1D24-25 ÒÓ·_·ĝôĞğ·õö÷øE_Úæ\M) xX
(1)𝑥2 = 81g
(2)(𝑥+ 1)3 = −27X
Þß(1)𝑥=± 9g
(2)𝑥= −4g
=àD1EĀ)„…Ž#𝑥)&FGH‡(]pHFĠġ—c(X
D2EĀ )„…Ž#𝑥)&F³o−27) F´ĕo𝑥+ 1)&F¢©o𝑥X
áĎďâãĐ> )„…kC"XĔ÷úû> DeüGH() fÁ𝑎Fýþ
ÿH(IJ𝑎) E){|d#)ùX
é#D1E#
𝑥2 = 81F 
𝑥=±
81
𝑥=± 9g
D2E#
(𝑥+ 1)3 = −27F 
𝑥+ 1 =
3 −27F
𝑥+ 1 = −3F
∴x = −4X
ð\ 2-2D24-25 ÒÓ·_·ĢôģĤĥ·ØÙEêë$(2𝑥−1)2−16 = 9) x )&      
X
Þß3 ‰−2
=àáâãĐ!"#$X
³;-ĕº(2𝑥−1)2 = 25F˜2𝑥−1 = 5‰2𝑥−1 = −5FĦ+œ†X
                                                                           
                                                                       
é##
∵(2𝑥−1)2−16 = 9
∴(2𝑥−1)2 = 25
∴2𝑥−1 = 5‰2𝑥−1 = −5
∴𝑥= 3‰𝑥= −2X
ÊÞß
−2‰ 3X
ð\ 2-3D24-25 ÒÓ·_·ħĨ·ØME_Ú$M x )&
(1)2𝑥2 = 50g
(2)
1
3(𝑥+ 1)2 = 3g
(3)(2−𝑥)3 = 64g
(4)64(𝑥+ 1)3−125 = 0
Þß(1)𝑥= 5‰𝑥= −5
(2)𝑥= 2‰𝑥= −4
(3)𝑥= −2
(4)𝑥=
1
4
=àáĎďâãĐ!"> ĩč()&FĔ÷úû#GH()k d#
)ùX
D1EĀ)„…#œ†g
D2EĀ)„…#œ†g
D3E!" )„…#œ†g
D4E!" )„…#œ†X
é#D1E#
2𝑥2 = 50
𝑥2 = 25F
#ĕ
𝑥= 5‰𝑥= −5g
D2E#
1
3(𝑥+ 1)2 = 3
(𝑥+ 1)2 = 9F
𝑥+ 1 = 3‰𝑥+ 1 = −3
#ĕ
𝑥= 2‰𝑥= −4g
D3E#
(2−𝑥)3 = 64
                                                                           
                                                                       
2−𝑥= 4F
#ĕ
𝑥= −2g
D4E#
64(𝑥+ 1)3−125 = 0
(𝑥+ 1)3 =
125
64 F
𝑥+ 1 =
5
4F
#ĕ
x = 1
4X
【题型3  由平方根、立方根求值】
• 3D24-25 ÒÓ·_·ÔÕĪī·õö÷øEČč|𝑎−6|? 𝑎+ 2𝑏—c(F𝑐+ 5) d 2F˜
𝑎−2𝑏−𝑐)     X
Þß ± 3
=àÅĎďâãĐc()„…F FF‡Éúûc„…d#ùX³!" 
—c()„…ĕo𝑎F𝑏F𝑐)&gĬĭ#ĕo𝑎−2𝑏−𝑐)&F´#œ†X
é##
∵|𝑎−6|? 𝑎+ 2𝑏—c(F
∴|𝑎−6| +
𝑎+ 2𝑏= 0F
∴𝑎−6 = 0F𝑎+ 2𝑏= 0F
#ĕ
𝑎= 6F𝑏= −3F
∵𝑐+ 5) d 2F
∴𝑐+ 5 = 8F
∴𝑐= 3
∴𝑎−2𝑏−𝑐= 6−2 × (−3)−3 = 9
∴𝑎−2𝑏−𝑐)d ± 3X
ÊÞß
 ± 3X
ð\ 3-1Čč x )pHd𝑎+ 3?2𝑎−15Fª2𝑏−1)+1d 3X
(1)𝑎,𝑏,𝑥)&g
(2)𝑎+ 𝑏−1) X
Þß(1)4F5F49
(2)2
=àáâãF+1k )„…XĔ÷úûL„…k6“d#)ùX
                                                                           
                                                                       
D1EĀ?+1)„…œ†ĕ𝑎F𝑏)&F´#𝑥)&œ†g
D2EĮ𝑎F𝑏)&Ĭĭ𝑎+ 𝑏−1MĦ+¶!" )„…œ†ĕÞßX
é#D1E#
 ∵𝑥)pHd𝑎+ 3?2𝑎−15Fª2𝑏−1)+1d 3F
∴𝑎+ 3 + 2𝑎−15 = 0F2𝑏−1 = 9F
#ĕ
𝑎= 4F𝑏= 5g
∴𝑥= (𝑎+ 3)2 = 72 = 49g
D2E#
 ∵𝑎= 4F𝑏= 5F
∴𝑎+ 𝑏−1 = 4 + 5−1 = 8
ð\ 3-2D24-25 ÒÓ·_·įİı¹·õö÷øEČč3𝑎+ 1)pH=Pd7−𝑚>2−2𝑚F9 + 𝑏)
 d 2X
(1)𝑚F𝑎F𝑏)&g
(2) 7𝑎−𝑏)X
Þß(1)𝑚= 3F𝑎= 5F𝑏= −1
(2) ±
6
=àÅâãF FĔ÷úûcčIJd#)ùg
D1EĀ„…F „…Úĕ7−𝑚+ 2−2𝑚= 0F9 + 𝑏= 23Fœ†o𝑚F𝑎F𝑏)&g
D2E³o7𝑎−𝑏)&F´!"„…oÞߜ†X
é#D1E#
sSĕF7−𝑚+ 2−2𝑚= 0 
#ĕ𝑚= 3        
Ê7−𝑚= 4,2−2𝑚= −4
∴3𝑎+ 1 = 42F
#ĕ𝑎= 5 
%S9 + 𝑏= 23F
#ĕ𝑏= −1g
D2E∵7𝑎−𝑏= 7 × 5−(−1) = 36F
36 = 6 F6 ) ±
6 F
ij^ 7𝑎−𝑏) ±
6X
ð\ 3-3D24-25 ÒÓ·_·ĴİĘĵ·ØÙEČč2𝑎+ 1)d ± 3.3𝑎+ 2𝑏−4) d−2F
(1)𝑎>𝑏)&g
                                                                           
                                                                       
(2)3𝑎+ 𝑏)
Þß(1)𝑎= 4F𝑏= −8
(2) ± 2
=àáĎďâãĐ F)„…FĀ F)„…#œ†X
D1E% F)„…†č2𝑎+ 1 = 9F3𝑎+ 2𝑏−4 = −8F­¶œ†o aFb )&X
D2EĶ aFb )&Ĭĭ3𝑎+ 𝑏F´œ†X
é#D1E#
 ∵2𝑎+ 1)d ± 3F3𝑎+ 2𝑏−4) d−2F
∴2𝑎+ 1 = 9F3𝑎+ 2𝑏−4 = −8F
∴𝑎= 4F
∴𝑏= −8X
D2E#
∵3𝑎+ 𝑏= 3 × 4−8 = 4F
∴3𝑎+ 2𝑏) ± 2X
【题型4  实数的运算】
• 4D24-25 ÒÓ·_·ķĸĹĺ·ØMEĮ 1F 2F 3F 6ƒeĻ\ļÚF–Ċ„(𝑚,𝑛)xyĽ𝑚ļ¸
¹đ»Ľ𝑛H(F˜(5,4)?(12,4)xy)p(qľd         
.
Þß0
=àÅĎďâãĐ(Ŀ)ðŀĊ«F'()*+FŁ‚(ĿðŀĊ«dùX
Ā()ļÚ¿†čFĽGļ
1 H(FĽĚļ 2 H(XĽłļ 3 H(FĽķļ 4 H(F…Ľ𝑚ļ]𝑚H
(F¸ĽGļº𝑚ļŃ]
1 + 2 + 3 + 4 + … + 𝑚=
𝑚(𝑚+1)
2
H(FĀ()ļÚ¿FŠķH(GHńŅF
ĀņSŇŁoĽ𝑚ļĽ𝑛H(ºňdʼnH(¶´Ħ+X
é##
Ā()ļÚ¿†čFĽGļ
1 H(FĽĚļ 2 H(XĽłļ 3 H(FĽķļ 4 H(F…
Ľ𝑚ļ]𝑚H(F¸ĽGļº𝑚ļŃ]
1 + 2 + 3 + 4 + … + 𝑚=
𝑚(𝑚+1)
2
H(FĀ()ļÚ¿FŠķH
(GHńŅF
                                                                           
                                                                       
(5,4)xyĽ 5 ļ¸¹đ»Ľ 4 H(d 2F
∵Ŋ 11 ļŃ]1 + 2 + 3 + 4 + … + 11 =
11×(11+1)
2
= 66 (H)(F
∴ (12,4)xyĽ 12 ļĽ 4 H(œĽ 70 H(F
∵70 ÷ 4 = 17⋯2F
∴ (12,4)xy)(d 2F
∴ (5,4)?(12,4)xy)p(qľd 2−2 = 0F
ÊÞß
0X
ð\ 4-1D2025 ŋÓ·Œ·āĂ·Ō÷øEĦ+
 16−(−1)2025−
3 27−|1−2|X
Þß3−2
=àáĎďâãĐ'()*+F³Ħ++1> FōĦ+›‘&>¦Fµ¶Ħ+¤¥
¿œ†ĕºÞßX
é##
Ŏ\ = 4−(−1)−3−( 2−1)
= 4 + 1−3−2 + 1
= 3−2X
ð\ 4-2D24-25 ŋÓ·Œ·ĝôŏĺ·ØÙEĊ„
[𝑚]xyUՐ𝑚)µ7;(F{𝑚}xy𝑚)8(<=F
[𝑚] + {𝑚} = 𝑚FLM𝑚'(X•e[3.14] = 3F{3.14} = 0.14F[3.14] + {3.14} = 3.14XĦ+
[−3]−{ 5}
=       
X
Þß−5
=àáâãĐʄ…_)'(*+F,-().+F-#ʄ…d#)ùX=P.Ħ−3> 5
ʼnpH;(qrF´Āʄ…ĕo[−3] = −2F{ 5} =
5−2Fpőc¥œ†ĕoÞßX
é##
∵−2 < −3 < −1F2 <
5 < 3F
∴[−3] = −2F{ 5} =
5−2F
∴[−3]−{ 5} = −2−( 5−2) = −5X
ÊÞß
−5X
ð\ 4-3D24-25 ÒÓ·Œ·ŒĴœò·ØMEŠH$±ö%–ŔŕŖŗŘzFřŚśĦĐGö*+$±e
Ļ
                                                                           
                                                                       
•e
ÍŜĭ x )&−1ÏFĦ+Ðü 3 < 4gĮŜĭ&ð(−1) +1ŝ0FĦ+Ðü 6 < 4g´ĮŜĭ
&ðĐ0 + 1 = 1FŞş*+FºĦ+ÐüU8Á 4FŠŜošÐüX
ŢŇâ_ÚĜX
(1)ÍŜĭ x )& 5F˜Ŝo y )&duţŤŢÚ\Ħ+X
(2)ÍťŦŜĭ x )& 1FŢȐĦ+ÛŧŨ`t$±*£¶ŠwŜo yX
Þß(1) 21
(2)4 t
=àáâãĐ'()*+F-#SFúûũĻM)*+¿˜d#)ùX
D1EĀũĻM)*+$±Ħ+œ†g
D2EĀũĻM)*+$±Ħ+FºÐü7Á‰fÁ 4 œŜoÐüŪX
é#D1EÍŜĭ x )& 5 ÏF
˜]5−(−2) = 7F7 × 3 = 21F
ª 21 > 4F
∴Ŝo y )&d 21X
D2EÍŜĭ x )& 1 ÏF
˜]1−(−2) = 3F3 × 3 = 9F 9 < 4FŞşĦ+g
ĽĚtŜĭ x )&1 + 1 = 2ÏF
˜]2−(−2) = 4F4 × 3 = 12F 12 < 4FŞşĦ+g
ĽłtŜĭ x )&2 + 1 = 3ÏF
˜]3−(−2) = 5F5 × 3 = 15F 15 < 4FŞşĦ+g
ĽķtŜĭ x )&3 + 1 = 4ÏF
˜]4−(−2) = 6F6 × 3 = 18F 18 > 4FŜo𝑦g
ij^Ũ 4 t$±*£¶ŠwŜo yX
【题型5  无理数的估算】
• 5D24-25 ÒÓ·_·ūŬŭŮ·ØÙE7ů­dŰ)śĦŚFœűdGŲ88)ųŴFŵŶmŷ“Ű
                                                                           
                                                                       
”XeĻF
𝐵𝐶
𝐴𝐵)&ËŸŹ9
5−1
2 F˜_Ú.+‡É)dD    E
AX0 <
5−1
2
<
2
5
BX
2
5 <
5−1
2
<
1
2
CX
1
2 <
5−1
2
< 1
DX1 <
5−1
2
< 2
ÞßC
=àáâãĐ,-().+Fáâã,-().ĦFUf\)6“F‡ÉźŻ)žŸd#á
)ùX%Sč4 < 5 < 9F˜2 <
5 < 3FĀUf\)6“Ħ+#œ†X
é##
%SčF 4 < 5 < 9F
∴2 <
5 < 3F
∴1 <
5−1 < 2F
∴
1
2 <
5−1
2
< 1X
Êç
CX
ð\ 5-1D24-25 ŋÓ·_·ĴżŽ·õö÷øEČč432 = 1849F442 = 1936F452 = 2025F462
= 2116X–𝑛;(ª𝑛<
2022 < 𝑛+ 1F˜𝑛)&D   E
AX43
BX44
CX45
DX46
ÞßB
=àáĎďâã,-().+FĔ÷úû,-().+d#)ùg³ĀŔM)(Ā.+ 2022
)78F¢©Ĝ†#X
é##
%S†č
1936 < 2022 < 2025
∴44 <
2022 < 45F
∴𝑛= 44g
Êç BX
ð\ 5-2D24-25 ĖÓ·Œ·ĴżŽ·ØME–𝑎<
23 < 𝑏Fª𝑎𝑏žş‡;(F˜𝑎+ 𝑏=        
Þß9
=àáâã'().+?789:)wſF³.+o 23)ƀ&žŸFĕo𝑎F𝑏)&F¢©†ĕoÐ
                                                                           
                                                                       
ÑXĀSo𝑎F𝑏)&d#)ùX
é##
∵16 < 23 < 25F
∴4 <
23 < 5F
∵𝑎F𝑏pHžş;(F
∴𝑎= 4F𝑏= 5F
∴𝑎+ 𝑏= 4 + 5 = 9X
ÊÞß
9X
ð\ 5-3D24-25 ÒÓ·_·ĢƁƂÔ·ØÙEČč𝑘<
17−1Fª𝑘‡;(F˜𝑘)&†^d           
DnoGHϠEX
Þß1DÞßUƃGE
=àáĎďâãĐ,-().+Uf\)#fčIJFúû,-().+¿d#)ùX
³.+ 17ĕƀ&žŸF´É„ k )ƀ&žŸF­¶ĀUf\#)„…œ†#ÞX
é##
∵16 <
17 <
25d;(g
∴3 <
17−1 < 4dƄā(g
∴𝑘≤3F
∵𝑘‡;(F
∴𝑘= 1‰ 2 ‰ 3X
ÊÞß
1DÞßUƃGEX
【拔尖篇】
【题型6  算术平方根的双重非负性】
• 6D24-25 ÒÓ·_·ĝƁƅƆ·õö÷øEČč aFb '(Fª𝑏=
𝑎−8−8−𝑎+25F˜
3 𝑎+
𝑏)
&D    EX
AX10
BX9
CX8
DX7
ÞßD
=àáâã+1z )ŕƇF FĀ+1)K(d45(ĕº𝑎= 8F¢©
ĕ𝑏= 25Fµ¶Ĭĭ
3 𝑎+
𝑏Ħ+œ†X
é##
∵𝑏=
𝑎−8−8−𝑎+25F𝑎−8 ≥0F8−𝑎≥0F
∴𝑎= 8F𝑏= 25F
                                                                           
                                                                       
∴
3 𝑎+
𝑏=
3 8 +
25 = 2 + 5 = 7F
Êç
DX
ð\ 6-1D25-26 ŋÓ·Œ·āĂ·ƈƉ÷øE–𝑎+
𝑎−3 = 3F 𝑎+ 3)&X
Þß 6
=àáĎďâã)dĚt\FĔ÷úûĚt\)6“d#)ùX
³ĮŎ\ŀ 𝑎−3 = 3−𝑎F´"Uf\Ř#œ†X
é##
ï𝑎+
𝑎−3 = 3F
ij^ 𝑎−3 = 3−𝑎F
ij^𝑎−3 ≥0,3−𝑎≥0F
ij^𝑎= 3F
ij^ 𝑎+ 3 =
3 + 3 =
6X
ð\ 6-2D24-25 ÒÓ·_·ñòñĺ·ØMEČč|𝑎−3| +
3 + 𝑏= 0F˜
𝑎
𝑏
2025
= D    E
AX1
BX−1
CX2
DX−2
ÞßB
=àáâ㛑&F+1F]-()¦F#)ùdo𝑎>𝑏)&X
Ā›‘&>+1)456F#ĕ𝑎>𝑏)&FĬĭĦ+œ†X
é##
|𝑎−3| +
3 + 𝑏= 0F|𝑎−3| ≥0F 3 + 𝑏≥0F
∴|𝑎−3| = 0F 3 + 𝑏= 0F
∴𝑎−3 = 0F3 + 𝑏= 0F
∴𝑎= 3F𝑏= −3F
∴
𝑎
𝑏= −1
∴
𝑎
𝑏
2025
= (−1)2025 = −1
Êç
BX
ð\ 6-3D2025 ŋÓ·_·ĝƁ·Ō÷øEČč4¨'( aFb êë|2𝑎−4| + |𝑏+ 2| +
(𝑎−4)𝑏2
+4 = 2𝑎F˜𝑎+ 𝑏=      
X
Þß2
=àáĎďâãĐ+1)456F›‘&)456FĮ\Ɗð[|2𝑎−4| + |𝑏+ 2| +
(𝑎−4)𝑏2
                                                                           
                                                                       
= 2𝑎−4F%+1)456F›‘&)456ĕo2𝑎−4 ≥0F´ŀƋ\Ɗ†ĕo|𝑏+ 2| +
(𝑎−4)𝑏2
= 0F´Ā+1)456F›‘&)456œ†ĕo𝑏= −2F𝑎= 4F¢©ĬĭĬ(\œ†ĕoÞ
ßX
é##
∵|2𝑎−4| + |𝑏+ 2| +
(𝑎−4)𝑏2 +4 = 2𝑎F
∴|2𝑎−4| + |𝑏+ 2| +
(𝑎−4)𝑏2 = 2𝑎−4F
∴2𝑎−4 ≥0F
∴2𝑎−4 + |𝑏+ 2| +
(𝑎−4)𝑏2 = 2𝑎−4F
∴|𝑏+ 2| +
(𝑎−4)𝑏2 = 0
∴𝑏+ 2 = 0F𝑎−4 = 0F
#ĕF𝑏= −2F𝑎= 4F
˜𝑎+ 𝑏= 4−2 = 2F
ÊÞß
2X
【题型7  实数的大小比较】
• 7D24-25 ÒÓ·_·ħĨƌƍĘ·ØME9:78
 11     3g−5     −2g
5+1
2     
3
2XDĆ“ > ”
“ = ”‰“ < ”EX
Þß     
>     
<     
>
=àáâãĐ'()789:FĔ÷úû^ŒčIJd#Þá)ùX
Ā'()789:¿#ޜ†X
é##
 ∵ 
11 >
9F
∴
11 > 3g
∵5 >
4 = 2F
∴−5 < −4F
∴−5 < −2g
∵5 >
4 = 2F
∴
5 +1 > 3F
∴
5+1
2
>
3
2g
                                                                           
                                                                       
ÊÞß
 > F < F > X
ð\ 7-1D24-25 ÒÓ·_·ÔÕƎò·ØME–𝑎=
3 7F𝑏=
5F𝑐= 2F˜ aFbFc )78ÀD    E
AX𝑏≤𝑐< 𝑎
BX𝑏< 𝑎< 𝑐
CX𝑎< 𝑐< 𝑏
DX𝑎< 𝑏< 𝑐
ÞßC
=àáâã'()789:F³!"ƏƐ¿.+ aFb )&F´9:78œ†X
é##
∵1 < 7 < 8F4 < 5 < 9F
∴
3 1 <
3 7 <
3 8F 4 <
5 <
9F
œ1 <
3 7 < 2F2 <
5 < 3F
∴1 < 𝑎< 2F2 < 𝑏< 3F
Ƒ𝑐= 2F
∴𝑎< 𝑐< 𝑏F
Êç
CX
ð\ 7-2D25-26 ŋÓ·Œ·āĂ·ƈƉ÷øE9: 65−2? 10 +2)78X
Þß 65−2 >
10 +2
=àÅâã,-().+>'()789:F†È.+pH,-()žŸF¢©É„pHĬ(\
)žŸF´¢£9:X
é##
ï8 <
65 < 9F3 <
10 < 4F
ij^6 <
65−2 < 7F5 <
10 +2 < 6F
ij^ 65−2 >
10 +2X
ð\ 7-3D25-26 ŋÓ·Œ·āĂ·ƈƉ÷øE9:_ÚæŘ()78
(1)
1
2?
10
6 g
(2)
6−1
2 ? 1X
Þß(1)
10
6 >
1
2
(2)
6−1
2
< 1
=àáâã'(9:78FĔ÷úû'(9:78)¿Fd#)ù
D1E.+¿9:78œ†g
D2E.+¿9:78œ†X
é#D1E#
∵10 >
9 = 3F
                                                                           
                                                                       
∴
10
6 >
3
6Fœ
10
6 >
1
2g
D2E∵4 <
6 <
9F
∴2 <
6 < 3F
∴1 <
6−1 < 2F
∴
6−1
2
<
2
2Fœ
6−1
2
< 1X
【题型8  求整数部分和小数部分】
• 8D24-25 ŋÓ·Œ·ĢôĘƒ·õö÷øE– 3);(<= aF8(<= bF4−3);(<=
cF8(<= dF˜
𝑏+𝑑
𝑎𝑐)&D   E
AX
1
2
BX
1
4
CX
3−1
2
DX
3+1
2
ÞßA
=àáâãĐ.+,-()78FȐ.+ 3ʼnpH;(qrF¸©É„𝑎= 1F𝑏=
3−1FÈ
.+4−3ʼnpH;(qrF¸©É„𝑐= 2F𝑑= 2−3F­¶Ķ abcd Ĭĭĕ(&X
é##
∵3);(<= aF8(<= bF1 <
3 < 2F
∴𝑎= 1F𝑏=
3−1F
∵1 <
3 < 2F
∴−2 < −3 < −1F
∴2 < 4−3 < 3F
∵4−3);(<= cF8(<= dF
∴𝑐= 2F𝑑= 4−3−2 = 2−3F
∴
𝑏+𝑑
𝑎𝑐=
3−1+2−3
1×2
=
1
2X
Êç
AX
ð\ 8-1D24-25 ÒÓ·_·įÕįò·ØME–6 +
5);(<=d𝑚F8(<=d𝑛F˜|𝑛−𝑚|D   E
AX 5−10
BX10−5
CX 5−6
DX6−5
ÞßB
=àáâã?,-();(<=])Ħ+F'()*+FƏƐ¿o𝑚,𝑛)&F´Ĭ&Ħ+œ†X
é##
∵4 <
5 <
9F
∴2 <
5 < 3F
                                                                           
                                                                       
∴8 < 6 +
5 < 9F
∴𝑚= 8F𝑛= 6 +
5−8 =
5−2F
∴|𝑛−𝑚| = | 5−2−8| = 10−5g
Êç BX
ð\ 8-2D24-25 ŋÓ·Œ·ĝôŏĺ·ØÙEĊ„
[𝑚]xyUՐ𝑚)µ7;(F{𝑚}xy𝑚)8(<=F
[𝑚] + {𝑚} = 𝑚FLM𝑚'(X•e[3.14] = 3F{3.14} = 0.14F[3.14] + {3.14} = 3.14XĦ+
[−3]−{ 5}
=       
X
Þß−5
=àáâãĐʄ…_)'(*+F,-().+F-#ʄ…d#)ùX=P.Ħ−3> 5
ʼnpH;(qrF´Āʄ…ĕo[−3] = −2F{ 5} =
5−2Fpőc¥œ†ĕoÞßX
é##
∵−2 < −3 < −1F2 <
5 < 3F
∴[−3] = −2F{ 5} =
5−2F
∴[−3]−{ 5} = −2−( 5−2) = −5X
ÊÞß
−5X
ð\ 8-3D24-25 ÒÓ·_·įÕĴƓ·õö÷øEČč'(8 +
13);(<=𝑎F8(<=d𝑚g'(
9−13);(<=𝑏F8(<=d𝑛X
(1)no𝑎F𝑚F𝑏F𝑛)&g
(2)12𝑚+ 12𝑛+
𝑎+ 𝑏−7)&)g
(3)
3 (𝑚+ 𝑛)2025 +(𝑎−2𝑏)+ 6)&X
Þß(1)𝑎= 11F𝑚=
13−3F𝑏= 5F𝑛= 4−13
(2) ± 3
(3)2
=àáâãĐ,-().+ FĔ÷úû^ŒčIJōƔƕ*"d#Å)ùX
D1E.+ĕo3 <
13 < 4F¸©†ĕ11 < 8 +
13 < 12F5 < 9−13 < 6FÐBSœ†ĕ#g
D2EĮD1EM𝑎= 11F𝑚=
13−3F𝑏= 5F𝑛= 4−13Ĭĭij\Ɗ¢£Ħ+F´ÐB)„…Ħ
+œ†ĕ#g
D3EĮD1EM𝑎= 11F𝑚=
13−3F𝑏= 5F𝑛= 4−13Ĭĭij\Ɗ¢£Ħ+F´ÐB )„…Ħ
                                                                           
                                                                       
+œ†ĕ#X
é#D1E#
∵9 < 13 < 16F
∴9 <
13 <
16Fœ3 <
13 < 4F
∴11 < 8 +
13 < 12F5 < 9−13 < 6F
∵'(8 +
13);(<=𝑎F8(<=d𝑚F'(9−13);(<=𝑏F8(<=d𝑛F
∴𝑎= 11F𝑚= 8 +
13−11 =
13−3F𝑏= 5F𝑛= 9−13−5 = 4−13g
D2E#
%D1E†ĕ
𝑎= 11F𝑚=
13−3F𝑏= 5F𝑛= 4−13F
12𝑚+ 12𝑛+
𝑎+ 𝑏−7
= 12 × ( 13−3) + 12 × (4−13) +
11 + 5−7
= 12 13−36 + 48−12 13 + 4−7
= 9F
∴12𝑚+ 12𝑛+
𝑎+ 𝑏−7)&) ±
9 =± 3g
D3E#
3 (𝑚+ 𝑛)2025 +(𝑎−2𝑏)+ 6
=
3 ( 13−3 + 4−13)2025 + (11−2 × 5) + 6
=
3 1 + 1 + 6
=
3 8
= 2X
【题型9  实数与数轴综合运用】
• 9D24-25 ÒÓ·_·ĢƁƖƗƘ·ØÙEƙĖƚ+1ƛMƜƝ1ƞ
“ƜƝæů¦Fō©K§qFœ
Ɵ”Fœ𝑐=
𝑎2 + 𝑏2D𝑎“Ɯ”F𝑏“Ɲ”F𝑐“Ɵ”E–“Ɯ”2F“Ɲ”3F˜“Ɵ”eĻijy(@Œ†xy
D    
E
AX𝐴
BX𝐵
CX𝐶
DX𝐷
ÞßC
=àáâã,-().+F'(?(@F-#SōÚĕ‡É)+\d#)ùXĀSÚ\
Ħ+¶.+L78F­¶É„L(@Œ)vƠœ†X
                                                                           
                                                                       
é##
–“Ɯ”2F“Ɲ”3F˜ 22 + 32 =
13F
∵9 < 13 < 16F
∴3 <
13 < 4F
˜“Ɵ”eĻijy(@Œ†xy𝐶F
Êç
CX
ð\ 9-1D24-25 ÒÓ·_·ơ×ƢÔ·ØMEeĻF‡[𝐴𝐵𝐶𝐷)¾ƣ 7FƤ A (@Œxy)(
−1F– E (@ŒD E  A )¹ƥEFª𝐴𝐷= 𝐴𝐸F˜ E ijxy)(          
X
Þß−7−1
=àáâã+1F'(?(@F³o‡[)ÃĞF¢©Āpr)ƦƧo E ijx
y)(ϠX
é##
∵‡[𝐴𝐵𝐶𝐷)¾ƣ 7F
∴𝐴𝐸= 𝐴𝐷=
7F
∵Ƥ A (@Œxy)(−1F
∴ E ijxy)(−7−1g
ÊÞß
−7−1X
ð\ 9-2D24-25 ŋÓ·_·ƨŭĴƩƪĴ·ØÙE'(𝑎F𝑏(@Œ)vƠeĻijyFŀƋ
|𝑎+ 1|−(𝑏−1)2
+
(𝑎−𝑏)2 =           
X
  
Þß2
=àáâãĐ'(?(@F!"(@ĕo−1 < 𝑎< 0F1 < 𝑏< 2F¢©ŀƋœ†X
é##
%(@Fĕ−1 < 𝑎< 0F1 < 𝑏< 2F
∴𝑎+ 1 > 0F𝑏−1 > 0F𝑎−𝑏< 0F
∴Ŏ\ = 𝑎+ 1−|𝑏−1| + |𝑎−𝑏|
                                                                           
                                                                       
= 𝑎+ 1−(𝑏−1)−(𝑎−𝑏)
= 𝑎+ 1−𝑏+ 1−𝑎+ 𝑏
= 2F
ÊÞß
2X
ð\ 9-3D24-25 ÒÓ·_·ĝôYƎ·ØMEeĻFȐƫÃĞ 1 )‡[FƬw‚É)Ķ 2xy(
@Œ𝐴1ƭFƮ𝐴1»ƥµË);(𝐵1F^𝐵1hƯF𝐴1𝐵1ưƱƫưhFƲ(@Á𝐴2FƮ𝐴2»ƥ
µË);(𝐵2F^𝐵2hƯF𝐴2𝐵2ưƱƫưhFƲ(@Á𝐴3FeÅŞşF˜𝐴2026𝐵2026)Ğ
          
X
Þß 2−1/−1 +
2
=àáâãĐ'()*+)Ċ«F'(?(@F³o B xy)(ĕº𝐴1𝐵1 = 2−2F˜𝐴2xy
)(4−2F´o𝐵2xy)(3F˜𝐴2𝐵2 =
2−1F­¶stxy𝐴3𝐵3 = 2−2F𝐴4𝐵4 =
2−1F𝐴5𝐵5
= 2−2g𝐴6𝐵6 =
2−1gœ†ŁºĊ«#X
é##
%SĕF𝐴1xy)( 2F
∵1 <
2 <
4F
∴1 <
2 < 2F
∴𝐵1xy)( 2F
F∴𝐴1𝐵1 = 2−2F˜𝐴2xy)(2 + 2−2 = 4−2F
∵1 <
2 < 2F
∴−2 < −2 < −1
∴2 < 4−2 < 3F
∴𝐵2xy)(3F
∴𝐴2𝐵2 = 3−(4−2) =
2−1F
²-†ĕ𝐴3𝐵3 = 2−2g
𝐴4𝐵4 =
2−1g
𝐴5𝐵5 = 2−2g
                                                                           
                                                                       
𝐴6𝐵6 =
2−1g
……F
^ŀƳ†ĕFÍ n ƴ(Ï𝐴𝑛𝐵𝑛= 2−2FÍ n Ƶ(Ï𝐴𝑛𝐵𝑛=
2−1F
∴𝐴2026𝐵2026 =
2−1
ÊÞß
 2−1X
【题型10  实数运算的应用】
• 10D24-25 ÒÓ·_·ĝôĞğ·õö÷øEƶŚƷ]GHƸƣ200cm3)ƹƺŎƻFƶŚƷƼďĮÿ
Hƹƺƽŀō3ĘƾƿzĘ)[ǀX
(1)–ĮŎƻ3ĘƾƿzGHň¾‡[ǁ8cm)ĞƸFĞƸň¾‡[)ÃĞX
(2)ƶŚƷĉĮŎƻƾƿzłH78c²)‡ƸƹƺFǂĄƄz¶ǃ_)ĵƻƸƣ8cm3FǂĄz)Š
GH8‡Ƹƹƺ)DŽĞX
Þß(1)5cm
(2)4cm
=àáâã+1F )C"F
D1EĀĞƸƸƣ)Ħ+\“ĞƸ)Ƹƣ = ň¾ƣ × ǁ”Ú$#œ†g
D2EĀ“‡ƸƸƣ)Ħ+¿^k3H8‡ƸƸƣ?ÄƸƣqr)À”Ú$#œ†g
-#+1 )„…d‡É#Þ)ùX
é#D1E#
śĞƸň¾‡[)ÃĞ𝑥cmF
sSFĕ
8𝑥2 = 200F
#ĕ
𝑥= 5‰𝑥= −5D5&UQBSFDždžEF
Þ
ĞƸň¾‡[)ÃĞ5cmg
D2E#
śŠGH8‡Ƹƹƺ)DŽĞ𝑦cmF
sSFĕ
3𝑦3 +8 = 200F
#ĕ
𝑦= 4F
Þ
ŠGH8‡Ƹƹƺ)DŽĞ4cmX
ð\ 10-1D24-25 ÒÓ·Œ·ŒĴ·ØMEeĻFGHň¾ưƱ3cm)LJƊ LjŷGlljNJXÍLJƊ‡
NjÏFLJ ljNJ)ǁnj16cmg™NjÏFċĵ<=)ǁnj4cmXLJ )ljNJ‡Ǎ™ê 2 HGǎ7)‡
ƸǏǐDπƀ 3FǏǐ)ǑnjUĦEX
                                                                           
                                                                       
(1)šLJƊ)ǏƣDLjêÏljNJ)ƸƣEduţ ǒǓŤ
(2)‡ƸǏǐ)DŽĞduţǒǓŤ
Þß(1)540cm3
(2)6cm
=àáâã]-()¯B*+GH() FÇǔkYZ`šƸ)ƸƣFǕǖSFĕo“LJ
Ɗ)Ǐƣ?²ňǁ(16 + 4)cm)hǗƸƣcf”d#)ùX
D1ELJƊ)Ǐƣ?²ňǁ(16 + 4)cm)hǗƸƣcfF%ņ#g
D2Eǘ³oLJ )ljNJ)ƸƣF­¶ĀLJ )ljNJ‡Ǎ™ê 2 HGǎ7)‡ƸǏǐ#œ†X
é#D1E#
𝑉= π𝑟2ℎ≈3 × 9 × (16 + 4) = 27 × 20 = 540(cm3)g
D2E#
ï𝑉溶液= π𝑟2ℎ≈3 × 32 × 16 = 432(cm3)X
ij^DŽĞ =
3 432 ÷ 2 = 6cmX
ð\ 10-2ǙGǚǛŒė¶FĐxǜăØsǝǞǟƄz(Ąą)8Ǡ>8ǡ²F(řŚǢ„
ǣÆǤǥŒ 2 ǦǧǨ‹d 16 ę)ǩǪǫǬǭġX¸ǩǪĥºǛ)ŠǮǯǧǰDZDz‹ 1.6 ęF%
Á(řŚdšǥ)dzǴFïŊǧ‹†^ű"GHǥǵDZ) 5 ęǥǶǷ‘ŠǮǯǧ)ÄǨ¥Ǹ 5
ęDǯǧÄǨUmǰDZDzF²GǯǧǹǺǻű"GHǶǷEXǼĀšǩǪĥ)ǽƕēFÍǯǧÄǨDU
mǰDZDzEê 30 ęÏF5 ę)ǥǶǷ†ǾǿGH 7 ę)ĥȀǶǷFœ†^ȁǯǧÄǨDUmǰDZDzE
¥Ǹ 7 ęÍ(řŚ²ÏĐ 2 ǦǩǪ‚Ȃ_ǧȃȄÏF8Ǡ²Û
“řŚFȅġ†^ǿGa_ǧ\F
ǽȆuȇ”Ţ²ġ=à8Ǡ²)_ǧ\FōĦ+oátÆǤÄDz"DǷmǰDZDzEµȈ†        
ęX
Þß25.2
=à=PĦ+pa_ǧ)\F9:ʼnGaÄDz"ȆȈœ†X
é#ĽGa_ǧ\ȉȊpǦǩǪ
BĦDz"
16 + 16 + 1.6−7 = 26.6ę
ĽĚa_ǧ\_pHǯǧFŠHǯǧȊGǦǩǪ
BĦDz"
(16 + 1.6−5) × 2 = 25.2ę
                                                                           
                                                                       
ÊçȋĽĚaȆȌ+FµȈDz" 25.2 ę
ÊÞß
25.2X
ěáâãĐ'(*+)'ȍC"F=€ȎÑd#)ùX
ð\ 10-3ǣ8O‚ȂKĈGƺĞ32mFȏ21m)Ğ[ċȐF
(1)ßG
eĻFĮÿƺċȐaŒȑȒFMrȓGŕȔȕ)8ȖF8Ȗ)¹Ãȗđ»Ē1mƬdĠ)»Ã
ȗX˜ÿƺȑȐ)¾ƣ _____ m2g
(2)ßĚ
ȓŬGHĞdȏ) 1.5 ȘF¾ƣ384m2)șȚțF–9Ȝ")șȚțďН23mº30mq
rFȏ13mº20mqrX ÿHșȚțw"J9ȜȝŤ ōÛŧ-%X
Þß(1)651
(2)wF-%Z#à
=àáâãĐĻ[)ĒF)„…fčIJX
D1E%SFȑȐ)Ğ¥81mFȏUðF渚ĕȑȐ)¾ƣg
D2Eśȏ𝑥mF˜Ğ1.5𝑥mFĀ¾ƣ\œ†ĕÁ x )$F%)„…œ†ĕ xF´‘ x )
&¢£.+F–êëSœ†FȞ˜U£X
é#D1E#
%SF8Ȗ)¹Ãȗđ»Ē1mƬdĠ)»Ãȗœ8Ȗ)ȏ1mF
˜ȑȐ)Ğ¥81mFȏUðF
¾ƣ21 × (32−1) = 651m2g
ÊÞß
651X
D2EwF-%e_
śȏ𝑥mF˜Ğ1.5𝑥mF
sS]
1.5𝑥2 = 384F 𝑥2 = 256
∵𝑥> 0F
∴𝑥= 16F1.5𝑥= 24
QBН23mº30mqrFȏ13mº20mqrF
∴ÿHșȚțw"J9ȜX