第二章 实数(举一反三讲义)全章题型归纳
2024
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12 ()*JKL ..................................................................................................................................10
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20 012*(^U# ..........................................................................................................................16
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`~π5b7π*),{ 2πa
π
2|}
_3`t-lmnoJ)s2*),{ 0.101 001 000 1…_
c 1 1bc 0)|p8-.)*J
)NO)-l}-.)*J)NOmno}-.)mO)*s2p
知识点7 实数的概念及分类
1. 34
7.)P-.)()p
2. O
()7
wO
_1`]9O
()O7.)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()t#)R*bc}a)R*bcbc()p()Q)R
*bb U.
知识点9 实数范围内的有关概念
¡
<=
¢£
y)
¤ a Q b ¥y)a¦𝑎+ 𝑏= 0
3*y)z−3
§)
¤ a Q b ¥§)a¦𝑎𝑏= 1
2 *§)z
1
2
k¨()*© 'zHI)aª|𝑎| ≥0
|𝑎|
𝑎(𝑎> 0)
0(𝑎= 0)
−𝑎(𝑎< 0)
© '
¥y)*
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
()Í 0aI()JÍ 0a()ÍbÎI()}
c()a© '
*()}
cI()a© '*I()J}«)R*
c()aÇÏ*)ÐKÅÏ*)
pÑÒaKL
()*JÓ7{uË
_1`ÔKL
()**Ja°·Õ]()*JÌp{KL 10Q 3 *Ja&Í( 10)
2 = 10
,32 = 9,10 > 9aÖ 10 > 3p
_2`#/,*Ë-.)*ר'ÂaÁKL
)*Jp
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_3`Ù
cÚ *-.)KLJÛaU#{uÜÝ
①𝑎> 𝑏≥0⇔𝑎>
𝑏p
②𝑎> 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!z 0
Cp 9*z±3
Dp−3z(−3)2*bc
ë2 1-1_24-25 ÞßÃu·ìíîáï·ðñòó`−
8
27*!z } 4*,êz
p
ë2 1-2_24-25 ÞßÃu·ôõ·öä÷ø`_1`ù
a
0.000001
0.001
1
1000
1000000
3 𝑎
_2`úûüý*XYùþ
①
3 3 = 1.442a¦
3 3000 = ______a
3 0.003 = ______}
②ÿĀ
3 343 = 7a¦
3 0.000343 = ______p
ë2 1-3¤ a b ¥y)ac d ¥I§)a¦ 𝑎2−𝑏2 +
3 𝑐𝑑=
.
【题型2 利用求平方根、立方根解方程】
£ 2_25-26 āß÷ôõ·ĂÂ÷ø` Í()𝑎,𝑏a]9bw+,“ ⊕”
𝑎⊕𝑏= 𝑎2−2𝑏pÍ𝑥*
%(𝑥+ 1) ⊕ 8 = 0*$
p
ë2 2-1_24-25 ÞßÃu·ăïĄą·ðñòó`uæĆ2A* xp
(1)𝑥2 = 81}
(2)(𝑥+ 1)3 = −27p
ë2 2-2_24-25 ÞßÃu·ćïĈĉĊ·äå`ċČ%(2𝑥−1)2−16 = 9* x *'
p
ë2 2-3_24-25 ÞßÃu·čĎ·äA`uæ%A x *'
(1)2𝑥2 = 50}
(2)
1
3(𝑥+ 1)2 = 3}
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(3)(2−𝑥)3 = 64}
(4)64(𝑥+ 1)3−125 = 0
【题型3 由平方根、立方根求值】
£ 3_24-25 ÞßÃu·àáďĐ·ðñòó`ÿĀ|𝑎−6|Q 𝑎+ 2𝑏¥y)a𝑐+ 5*!z 2a¦
𝑎−2𝑏−𝑐* p
ë2 3-1ÿĀ x *
cz𝑎+ 3Q2𝑎−15a¸2𝑏−1*,êz 3p
(1)𝑎,𝑏,𝑥*'}
(2)𝑎+ 𝑏−1*!p
ë2 3-2_24-25 ÞßÃu·đĒēÅ·ðñòó`ÿĀ3𝑎+ 1*
cOiz7−𝑚P2−2𝑚a9 + 𝑏*
!z 2p
(1)𝑚a𝑎a𝑏*'}
(2) 7𝑎−𝑏*p
ë2 3-3_24-25 ÞßÃu·ĔĒ\ĕ·äå`ÿĀ2𝑎+ 1*z ± 3.3𝑎+ 2𝑏−4*!z−2a
(1)𝑎P𝑏*'}
(2)3𝑎+ 𝑏*
【题型4 实数的运算】
£ 4_24-25 ÞßÃu·ĖėĘę·äA`Ě 1a 2a 3a 6{ě2Ĝæa¤X](𝑚,𝑛)ĝ𝑚ĜÄ
ÅĞÇĝ𝑛c)a¦(5,4)Q(12,4)*
)ğz
.
ë2 4-1_2025 Ġß÷ôõ·ġòó`Ģ,
16−(−1)2025−
3 27−|1−2|p
ë2 4-2_24-25 Ġß÷ăïģę·äå`X]
[𝑚]mĤ𝑚*@M)a{𝑚}𝑚*J)NOa
[𝑚] + {𝑚} = 𝑚a6A𝑚()p£{[3.14] = 3a{3.14} = 0.14a[3.14] + {3.14} = 3.14pĢ,
[−3]−{ 5}
=
p
ë2 4-3_24-25 Þß÷ĥĔĦí·äA`c%¾ñ&¤ħ:ĨĩĪaīĬĢĭbñ+,%¾{
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ě
£{
ÙĮį x *'−1ÛaĢ,Üİ 3 < 4}ĚĮį'ë(−1) +1ı0aĢ,Üİ 6 < 4}ÁĚĮį
'ëĭ0 + 1 = 1aIJij+,aÆĢ,ÜİmJÍ 4aĴĮĵÜİp
ĶķĸuæĹp
(1)ÙĮį x *' 5a¦Į y *'zĺĻĶæ2Ģ,p
(2)ÙļĽĮį x *' 1aĶÔĢ,çľĿv1%¾+±ÂĴĮ yp
【题型5 无理数的估算】
£ 5_24-25 ÞßÃu·ŀŁłŃ·äå`ńºzŅ*ĬĢaªņzbŇJJ*ňʼnaŊŋŌ“Ņ
”p{ěa
𝐵𝐶
𝐴𝐵*'×ōŎK
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
ë2 5-1_24-25 ĠßÃu·ĔŏŐő·ðñòó`ÿĀ432 = 1849a442 = 1936a452 = 2025a462
= 2116p¤𝑛M)¸𝑛<
2022 < 𝑛+ 1a¦𝑛*'_ `
Ap43
Bp44
Cp45
Dp46
ë2 5-2_24-25 āß÷ĔŏŐő·äA`¤𝑎<
23 < 𝑏a¸𝑎 𝑏ŒijM)a¦𝑎+ 𝑏=
ë2 5-3_24-25 ÞßÃu·ćœà·äå`ÿĀ𝑘<
17−1a¸𝑘M)a¦𝑘*'tz
_bcª`p
【题型6 二次根式的概念及其有意义的条件】
£ 6_24-25 ĠßÃu·àáŔí·ðñòó`ÿĀ 3𝑥−6 +
6−3𝑥+𝑦= 2025a¦ 2025𝑥𝑦*'_`
Ap2025 3
Bp2025 2
Cp2024
Dp2025
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ë2 6-1_24-25 ĠßÃu·đááŕ·ðñòó`uæĆ2Aab]z012*z_ŖŖ`
Ap −2
Bp
3 3
Cp 5
Dp 𝑎+ 1
ë2 6-2_24-25 ĠßÃu·ăïŗę·fĸŘ`ÿĀ
𝑦=
𝑥−4+ 4−𝑥
3
−2a¦3𝑥+ 4𝑦=
p
ë2 6-3_24-25 ĠßÃu·ăïŗę·fĸŘ`¤𝑥 𝑦()a¸ċČ2 4−𝑥+ |𝑦| = 𝑥−4a¦ 𝑥−𝑦*
,ê
p
【题型7 利用二次根式的性质化简】
£ 7_24-25 ĠßÃu·ŀŁłŃ·ðñòó`řO2𝑎−
1
𝑎a Ò*ŚW a ś° ®*Üİz_ `
Ap 𝑎
Bp −𝑎
Cp−𝑎
Dp−−𝑎
ë2 7-1_24-25 āß÷ĖėŜà·ðñòó`¤𝑥𝑦> 0a¦ −𝑥2𝑦>?_ `
Ap𝑥−𝑦
Bp−𝑥𝑦
Cp−𝑥𝑦
Dp−𝑥−𝑦
ë2 7-2¤𝑚< 0a¦ř2ŝ𝑚𝑥 ÒÊ*Ş2śÆ ÉÂ*2ŝUz_ `
Ap−𝑚2𝑥
Bp −𝑚2𝑥
Cp (−𝑚)2𝑥
Dp−−𝑚𝑥
ë2 7-3¤ 1575𝑛zM)abcM)𝑛*'
p
【题型8 最简二次根式】
£ 8_24-25 Ġß÷şĒŠï·ðñòó`ÿĀ 𝑥2 + 1z@?012a¸Q 5tTša¦𝑥*'
_ `
Ap2
Bp−4
Cp4
Dp ± 2
ë2 8-1_24-25 Ġß÷Ţōţ·äå`uæ012Aaz@?012*z_ `
Ap 18
Bp 63𝑎
Cp 𝑎2−2𝑎+ 4
Dp
𝑎
3
ë2 8-2Ě2ŝ 35−𝑎_a M)`>@?012ÂatQ 8TšpbcjT:; a *
'
p
ë2 8-3ÿĀ𝐴= 2 2𝑥+ 1a𝐵= 3 𝑥+ 3a𝐶=
10𝑥+ 3𝑦a6A AaB @?012a¸
𝐴+ 𝐵= 𝐶a¦2𝑦−𝑥*'
p
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【题型9 求二次根式中的参数值】
£ 9¤(2−7)
2
= 𝑎−𝑏7a6A𝑎a𝑏ŤM)a¦𝑏−𝑎*'
p
ë2 9-1012 2𝑏+ 1Q 𝑎−1 *P 0a¦𝑎+ 𝑏*'
p
ë2 9-2_2025 āßÃu·ć·ġòó`¤
𝑚
2 = 2 2a¦M)𝑚*'
p
ë2 9-3ÿĀ 10−𝑛zM)a¦M)𝑛ť7'Ŧ7
caŧŨ*P
p
【题型10 二次根式的运算或化简求值】
£ 10ÿĀ𝑥=
2 +1a𝑦=
2−1auæĆ2*'p
(1)𝑥2 + 𝑦2}
(2)
𝑥
𝑦+
𝑦
𝑥p
ë2 10-1Ģ,
(1) 3 +3
1
3−12}
(2) 45 ÷
27 ×
3
5}
(3) 18 ÷
2−32 +
3 ×
6}
(4)(3 +
3)(3−3) + (1 +
7)
2
p
ë2 10-2_24-25 Ġß÷Ėėũí·ðñòó`ÿĀ𝑥= 2−3aŪ)2(7 +
3)𝑥2 + (2 +
3)𝑥+
3
*'p
ë2 10-3ÿĀ 𝑥+ 𝑦= −5a𝑥𝑦= 4a𝑥
𝑦
𝑥+𝑦
𝑥
𝑦*'p
【拔尖篇】
【题型11 二次根式的双重非负性】
£ 11_24-25 ÞßÃu·ăūŬ·ðñòó`ÿĀ aab ()a¸𝑏=
𝑎−8−8−𝑎+25a¦
3 𝑎+
𝑏
*'_ `p
Ap10
Bp9
Cp8
Dp7
ë2 11-1_25-26 Ġß÷ôõ·ŭŮòó`¤𝑎+
𝑎−3 = 3a 𝑎+ 3*'p
ë2 11-2_24-25 ÞßÃu·ìíìę·äA`ÿĀ|𝑎−3| +
3 + 𝑏= 0a¦
𝑎
𝑏
2025
= _ `
Ap1
Bp−1
Cp2
Dp−2
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ë2 11-3_2025 ĠßÃu·ă·ġòó`ÿĀH¶() aab ċČ|2a−4| + |b + 2| +
(a−4)b2
+4 = 2𝑎a¦𝑎+ 𝑏=
p
【题型12 实数的大小比较】
£ 12_24-25 ÞßÃu·čĎůŢ\·äA`KLJ
11 3}−5 −2}
5+1
2
3
2p_ù“ > ”
“ = ”C“ < ”`p
ë2 12-1_24-25 ÞßÃu·àáŔí·äA`¤𝑎=
3 7a𝑏=
5a𝑐= 2a¦ aabac *JÌ
_ `
Ap𝑏≤𝑐< 𝑎
Bp𝑏< 𝑎< 𝑐
Cp𝑎< 𝑐< 𝑏
Dp𝑎< 𝑏< 𝑐
ë2 12-2_25-26 Ġß÷ôõ·ŭŮòó`KL 65−2Q 10 +2*Jp
ë2 12-3_25-26 Ġß÷ôõ·ŭŮòó`KLuæĆĪ)*J
(1)
1
2Q
10
6 }
(2)
6−1
2 Q 1p
【题型13 求整数部分和小数部分】
£ 13_24-25 Ġß÷ćï\Ű·ðñòó`¤ 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
ë2 13-1_24-25 ÞßÃu·đáđí·äA`¤6 +
5*M)NOz𝑚aJ)NOz𝑛a¦|𝑛−𝑚|_ `
Ap 5−10
Bp10−5
Cp 5−6
Dp6−5
ë2 13-2_24-25 Ġß÷ăïģę·äå`X]
[𝑚]mĤ𝑚*@M)a{𝑚}𝑚*J)N
Oa[𝑚] + {𝑚} = 𝑚a6A𝑚()p£{[3.14] = 3a{3.14} = 0.14a[3.14] + {3.14} = 3.14pĢ,
[−3]−{ 5} =
p
ë2 13-3_24-25 ÞßÃu·đáĔű·ðñòó`ÿĀ()8 +
13*M)NO𝑎aJ)NOz𝑚}(
)9−13*M)NO𝑏aJ)NOz𝑛p
(1)𝑎a𝑚a𝑏a𝑛*'}
(2)12𝑚+ 12𝑛+
𝑎+ 𝑏−7*'*}
(3)
3 (𝑚+ 𝑛)2025 +(𝑎−2𝑏)+ 6*'p
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【题型14 实数与数轴综合运用】
£ 14_24-25 ÞßÃu·ćŲųŴ·äå`ŵāŶ,êŷAŸŹêź
“ŸŹĆń´aš·fµaª
Ż”aª𝑐=
𝑎2 + 𝑏2_𝑎“Ÿ”a𝑏“Ź”a𝑐“Ż”`¤“Ÿ”2a“Ź”3a¦“Ż”«{ěť)R
«_
`
Ap𝐴
Bp𝐵
Cp𝐶
Dp𝐷
ë2 14-1_24-25 ÞßÃu·żãŽà·äA`{ěas𝐴𝐵𝐶𝐷*Êž 7aſ A «)R*)
−1a¤ E «)R_ E « A *Åƀ`a¸𝐴𝐷= 𝐴𝐸a¦ E ť*)
p
ë2 14-2_24-25 ĠßÃu·ƁłĔƂƃĔ·äå`()𝑎a𝑏«)R*Ƅ{ěťa>?
|𝑎+ 1|−
(𝑏−1)2 +
(𝑎−𝑏)2 =
p
ë2 14-3_24-25 ÞßÃu·ăïqŔ·äA`{ěaÔƅÏĄ 1 *saƆÕ*ř 2«
)R𝐴1Ƈaƈ𝐴1Çƀ@×*M)𝐵1at𝐵1~Ɖa𝐴1𝐵1ƊƋƅƊ~aƌ)RÍ𝐴2aƈ𝐴2Ç
ƀ@×*M)𝐵2at𝐵2~Ɖa𝐴2𝐵2ƊƋƅƊ~aƌ)RÍ𝐴3a{ÑIJija¦𝐴2026𝐵2026*Ą
p
【题型15 实数运算的应用】
£ 15_24-25 ÞßÃu·ăïĄą·ðñòó`ƍƎ7bcƏž200cm3*ƐƑƒƓaƍƎƔƕĚƖ
cƐƑƗ>šG\Ƙƙ\*sƚp
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(1)¤ĚƒƓG\ƘƙbcƛÊs Ɯ8cm*ĄƏaĄƏƛÊs*ÏĄp
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ë2 15-1_24-25 Þß÷ĥĔ·äA`{ěabcƛÊƊƋ3cm*Ƣŝ®ƣŌbƤƥpÙƢŝ
ƦÛaƢ®Ƥƥ*ƜƧ16cm}§ƦÛaþĕNO*ƜƧ4cmpƢ®*Ƥƥƨ§ċ 2 cbƩ*
Əƪƫ_πƬ 3aƪƫ*ƭƧmĢ`p
(1)ĵƢŝ*ƪž_ƣċÛƤƥ*Əž`zĺ!ƮƯĻ
(2)Əƪƫ*ơĄzĺƮƯĻ
ë2 15-2ưbƱƲĂÂaĭƳ«öäƴƵƶƟ)÷ø*JƷPJƸ¿a)īƹ]
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_LJƾÐƿmLjljNJa¿bLJƾǑǒǓņ#bcǎǏ`pǔúĵǁǂĊ*ǕǖǗaÙLJƾÐƿ_m
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Ǯp¦ƖƑǨǦ*Êž _____ m2}
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aǥ«13mÆ20mp ƖcǰDZDz#eKdzǴĻ šçľ.&p
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【题型16 复合二次根式的化简】
£ 16ǵǶuæĆ2
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……pĶ+#t*Ë>?
7 + 2 10 =
p
ë2 16-1_24-25 āßÃu·àáůí·fĸŘ`|2−3| +
4−2 3 = _ `
Ap2 3−3
Bp2 3−1
Cp3
Dp1
ë2 16-2Ĭ𝑎−13 +
5−3−5*J)NOa¦𝑎= _ŖŖ`
Ap
7
5
Bp
2
2
Cp 2−1
Dp 3−1
ë2 16-3_24-25 Ġß÷ĥĔǷí·äå`>? 23−6 10 + 4 3−2 2*Üİz_ `
Ap3 +
2
Bp3−2 2
Cp3 + 2 2
Dp3−2
【题型17 利用分母有理化求值】
£ 17_24-25 Ġß÷Ėė·äA`ǸǹuÊĢ,%
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
Ř
(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
ë2 17-1Ô»Ù+,aĚOWA*012>7.2*%aOW7.>p
{
1
2 =
2
2 a
1
3−2 =
1×( 3+ 2)
( 3−2)_ 3+ 2`
=
3 +
2Ɩ+,OW7.>p
(1)Ěuæ012OW.>
2−3
12 = ___________a
𝑥−𝑦
𝑥−𝑦= ___________
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(2)Ǻ ǻ
Ǽ>?
𝑥−𝑦
𝑥+ 𝑦Ûa
wm¿*$ǽ%p
Ǻ
𝑥−𝑦
𝑥+ 𝑦=
(𝑥−𝑦)
𝑥−𝑦
𝑥+ 𝑦
𝑥−𝑦=
(𝑥−𝑦)
𝑥−𝑦
𝑥−𝑦
=
𝑥−𝑦}
ǻ
𝑥−𝑦
𝑥+ 𝑦=
( 𝑥)
2−( 𝑦)2
𝑥+ 𝑦
=
𝑥+ 𝑦
𝑥−𝑦
𝑥+ 𝑦
=
𝑥−𝑦
ĶûǾǿǸǹǺ ǻ
Ǽ*$%a Ǻ ǻ
Ǽ*$ǽ÷Ȁȁ_ŖŖ`
ApǺ ǻè BpǺèǻ CpǺ ǻô DpǺ ǻôè
(3)ÿĀ7.) a b ċČ
𝑎
2+1 +
𝑏
2 = −1 + 2 2a a b *'p
ë2 17-2_24-25 Ġß÷ćȂ]·äA`ǸǹȃƓ
«°±012>?ÛaǟŨ7ÛNjȄ{
3
5a
2
3a
2
3+1ƖƩb*2ŝa6(ǟŨÓtĚ6°bȅ>
?
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 =
3−1ptƖw>?*ȅȆd
eOW7.>p
$ƹĹ
(1)ȇĐÊ*$%a>?
1
7−6 = __________p
(2)Ģ,
1
2+1 +
1
3+ 2 +
1
4+ 3 + ⋯+
1
2025+ 2024 × ( 2025 + 1)p
(3)ÿĀ𝑎=
1
3+ 2a𝑏=
1
3−2a𝑎2 + 𝑏2*'p
ë2 17-3_24-25 Ġß÷GȈ·äA`£{
2+1
2−1 =
( 2+1)( 2+1)
( 2−1)( 2+1) = 2 2 +3pȉƖƩÔOŝ OW
¿´bc2ŝřOWA* >ȊCȋř A*OW>ȊadeOW7.>p7uæÜÝ
①¤ a z 3*J)NOa¦
3
𝑎*'3 3 +6}
②KLJ
2024−2023 <
2023−2022}
③ës
1
3+ 2+1 =
2+ 2−6
4
}
④Ģ,
2
3+ 3 +
2
5 3+3 5 +
2
7 5+5 7 +⋯+
2
99 97+97 99 = 1−
11
33 }
⑤ÿĀ𝑎=
𝑚+1−𝑚
𝑚+1+ 𝑚a𝑏=
𝑚+1+ 𝑚
𝑚+1−𝑚a¸𝑎2 +1926𝑎𝑏+ 𝑏2 = 2024a¦ť7*M) m *P−1p
tÜÝÕ*z_ `
Ap①③④⑤
Bp②③④
Cp②③⑤
Dp②③④⑤
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【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+ 32Üİz_ `
Ap1
Bp
2 2
3
Cp2 2
Dp4 2
ë2 18-1_24-25 āßÃu·àáȌƼ·äå`ǵǶuæ|2
①3 + 2 2 = ( 2 + 1)
2
}②5 + 2 6 =
( 3 +
2)
2
}③7 + 2 12 = (2 +
3)
2
}…}ĶútXYaĝ 9 c|2
p
ë2 18-2Ô“&ȍȎÆbȏ”*ËZ[uÊ012*+,XY
ȍ£ 1
1 +
1
3 =
3+1
3 =
4 ×
1
3 = 2
1
3}ȍ£ 2
2 +
1
4 =
8+1
4 =
9 ×
1
4 = 3
1
4}ȍ£ 3
3 +
1
5 =
15+1
5
=
16 ×
1
5 = 4
1
5……U#üý*+
,XY 2024 +
1
2026 ×
4052*'_ `
Ap2024
Bp2025 2
Cp2023
Dp2023 2
ë2 18-3_24-25 ĠßÃu·àáŔí·äA`ÀǵǶuæ|2aÁȐǽĹ
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}
…
Ķû"#üý*XYaĢ,
1 +
1
12 +
1
22 +
1 +
1
22 +
1
32 +
1 +
1
32 +
1
42 +⋯+
1 +
1
20222 +
1
20232−2024 =
p
【题型19 二次根式的新定义】
£ 19_24-25 ĠßÃu·ĥĔȑȒ·ðñòó`]9
ǟŨĚ( 𝑎+
𝑏)Q( 𝑎−𝑏)b “ ȓ2”pŞ
( 𝑎+
𝑏)( 𝑎−𝑏) = ( 𝑎)
2
−( 𝑏)
2
= 𝑎−𝑏at7Ȕ*Ȋȕ aťt7bĹtÔȖƙ“ ȓ2”
$ƹp£{
ÿĀ 18−𝑥−11−𝑥= 1a 18−𝑥+
11−𝑥*'atƖƩ$ǽ
Ş( 18−𝑥−11−𝑥) × ( 18−𝑥+
11−𝑥) = ( 18−𝑥)
2
−( 11−𝑥)
2
= 18−𝑥−11 + 𝑥 = 7aȗŞ 18−𝑥−
11−𝑥= 1aťt 18−𝑥+
11−𝑥= 7p
_1`ÿĀ
20−𝑥+
4−𝑥= 8a¦𝑥*'z
}
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_2`Ģ,
1
3 1+ 3 +
1
5 3+3 5 +
1
7 5+5 7 +⋯+
1
2023 2021+2021 2023 =
ë2 19-1_24-25 ĠßÃu·àáŔí·äA` Ík8my|*
c()𝑎,𝑏a]9+,※{u
Ù𝑎< 𝑏
Ûa𝑎※𝑏= 2 𝑎+
𝑏aÙ𝑎≥𝑏Ûa𝑎※𝑏= 2 𝑎−𝑏a£{5※2 = 2 5−2aȘX]aĢ, 3※2 −
8※12 *Üİ_ `
Ap4 3−5 2
Bp−5 2
Cp4 3−3 2
Dp3 2
ë2 19-2_24-25 ĠßÃu·ĖėĘę·ðñòó`]9
Ş( 𝑎+
𝑏)( 𝑎−𝑏) = ( 𝑎)
2
−( 𝑏)
2
= 𝑎−𝑏at7Ȕ*Ȋȕ aǟŨ( 𝑎+
𝑏)Q( 𝑎−𝑏)b “ ȓ2”p¤ 18−𝑥−11−𝑥= 1a¦
18−𝑥+
11−𝑥= p
ë2 19-3 Ík8() mana¤]9\+,𝑚⊗𝑛=
𝑚−𝑛(𝑚≥𝑛),
𝑚+
𝑛(𝑚< 𝑛) aǛƝcçË
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
tçËAÕ*c)z_ `
Ap0 c
Bp1 c
Cp2 c
Dp3 c
【题型20 二次根式的实际应用】
£ 20{ěaĚbƐșȚțytbcĄ
8𝜋aǥ
2𝜋*Ȝsa¤ĚƖƐșȝfG\Ț
țybc~sa¦ĵ~*Êžz _ `
Ap12π
Bp18π
Cp24π
Dp36π
ë2 20-1{ěa«sȞŇAƦƄ
cJsaÿĀ
cJs*ÊžOi𝑆1 = 18a𝑆2
= 12aGȟNOzbcsa6Êž 2a¦þȠNO*Êž
p
ë2 20-2_24-25 Ġß÷àĒȡA·äA`üȢƌÔȣÖÂaƌȤȥǶÔqúȦȧÂȧȨȩ*Ȫ
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ȫ/ĢȧȬ±ȭ*ȮƧať#*Ŀȯ2z𝑣= 16 𝑑𝑓a6A𝑣ȧȮ_ƾ
km h`a𝑑ȦȧÂȧ
Ȩȩ*Ȫȫ_ƾ
m`a𝑓ȰȱŞ)p«ƺ1ƌÔȣÖȲȳAaȴȵ𝑑= 20ma𝑓= 1.2a¦ȶȣȷ
ȧ*ȧȮȸzĺĻ_ 6 ≈2.45aÜİȹÕÆ1km/h`
ë2 20-3_24-25 Ġß÷ĔĒïȺ·äå`ƜþȻȼzbwmȽľ*Ⱦȿ±aúɀ[aȼɁÄȫǦ
Ê h Ư*ƜƇń&ɂuaɂÆǦÊ*Û taĿ(ȯaüý𝑡=
ℎ
5(mĸɃɄɅ*Ɇɇ)p
(1)ȼƏÄ40m*ƜþɂÆǦÊ*Û}
(2)ÿĀÄƜþɈɂ*ȼƏťÚɉ(ƾ
J) = 10 × ȼƏ=ɉ(kg) × ƜƧ(m)abɊ=ɉ0.2kg*ɋɌĿ
3sɂ«ǦaƖɊɋɌ«uɂ%AťÚɉ7Ļ
(3)«_2`*ÜİAaûȵÆɍɎɏĻ(
ɐɑ-ɒɓǼƏǑɔƕ65J*ɉ)
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