第二十章二次根式(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(沪教)

第二十章  二次根式（举一反三讲义）全章题型归纳
 2024
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 1   !"#$%&' () ....................................................................................................4
 2  *+ ,-./ ....................................................................................................................5
 3  0/ ........................................................................................................................................7
 4  12 345 ........................................................................................................................9
 5   678./15 ..............................................................................................................10
9: .................................................................................................................................................................13
 6  ;< ./ ..........................................................................................................................13
 7  *+=>%?.15 ..........................................................................................................................15
 8   @ABC ..........................................................................................................................21
 9   DE' ..............................................................................................................................24
 10  FGH+ ..........................................................................................................................27
知识点1  二次根式的概念
1.E'
IJKLMNOPQ 𝑎R𝒂≥𝟎S TUVL“
”UV	La UVWXY4Z
2.[\
]^_`abcd()
R1Se	“
”fR2SWXY4]^ghi4RWXY
4jkg4lmjkge%l> TSZ
知识点2  二次根式有无意义的条件
当𝑎≥0时，二次根式𝑎有意义；
当𝑎< 0`L 𝑎无意义．
nQ
op𝑥2 +2𝑥+ 1 = (𝑥+ 1)2 ≥0Lqk 𝑥2 + 2𝑥+ 1r%&'Z
知识点3  二次根式的性质
1. 𝑎s%tuhi,
 被开方数𝑎≥0L
二次根式𝑎≥0．
2.( 𝑎)2 = 𝑎（𝑎≥0）LvIdhi4 7wxY xYyz{|}LQ( 5)
2 = 5Z
3. 𝑎2 = |𝑎| =
𝑎(𝑎≥0)L
−𝑎(𝑎< 0)LvId~&F4 xY 7wxYyz{|} €5Z
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Q ( ± 11)2 = | ± 11| = 11Z
[\
( 𝑎)2(𝑎≥0) 𝑎2 ‚ƒ
运算结果
( 𝑎)2 = 𝑎
𝑎2 =
𝑎(𝑎≥0)
−𝑎(𝑎< 0)
a 的取值
𝑎≥0
任意实数
作用
①用来去根号，化简二次根式；
②可用𝑎= ( 𝑎)2将任意一个非负实数
写成一个数的平方的形式
①+„…	L./f
②†	‡ hioxYˆ‰Š
	‹. nQ
Œ𝑎≥0, 2𝑎=
2𝑎2
知识点4  二次根式的乘法
1.  Ž
cd7wxY Lyz{NWXY4  7wxYLv 𝑎∙
𝑏=
𝑎𝑏(𝑎≥0，𝑏≥0)Z
nQ
 2 ∙
32 =
2 × 32 =
64 = 8Z
2.  Ž [\
R1S Žj‘’Š“d”Ž 67Lv 𝑎∙
𝑏∙
𝑐=
𝑎𝑏𝑐(𝑎≥0，𝑏≥0，𝑐≥0)Z
R2S•–—%˜4`L™š›œŽL†	– ˜4”ŽLp ˜4Lv𝑚𝑎∙𝑛𝑏= 𝑚𝑛𝑎𝑏
(𝑎≥0，𝑏≥0)Z
知识点5  积的算术平方根
1.  7wxYyz2ždo7wxY Lv 𝑎𝑏=
𝑎∙
𝑏(𝑎≥0，𝑏≥0)Z
6+Ÿ`LWXY4]^ ¡¢Ž PZ
2. £jk‘’Š“dhi4  7wxY 67Lv 𝑎𝑏𝑐=
𝑎∙
𝑏∙
𝑐(𝑎≥0，𝑏≥0，𝑐≥0)Z
3. H+
./L¤†WXY4¥¦o4=§8o=§L¨*+ 𝑎𝑏=
𝑎∙
𝑏(𝑎≥0，𝑏≥0)
𝑎2 = 𝑎(𝑎≥0)L† X©ªY o48oXŠ	‡Z
知识点6  二次根式的除法
1.  «
cd7wxY ¬Lyz{NWXY4 ¬ 7wxYLv
𝑎
𝑏=
𝑎
𝑏(𝑎≥0，𝑏> 0)Z
2.  «jk‘’Š“d”« 67Lv 𝑎÷
𝑏÷
𝑐=
𝑎÷ 𝑏÷ 𝑐(𝑎≥0，𝑏> 0
，𝑐> 0)Z
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3. –—%˜4`Lj™š›œ­›œ”« LO˜4WXY4=ƒ”«p oL
v𝑚𝑎÷ 𝑛𝑏= (𝑚÷ 𝑛) ∙
𝑎÷ 𝑏Z
知识点7  商的算术平方根
¬ 7wxYyz¬2ždo7wxY ¬Lv
𝑎
𝑏=
𝑎
𝑏(𝑎≥0，𝑏> 0)Z
知识点8  最简二次根式
1. WXY4®e=>L¯°WXY42®e X©ªY o4R8oSL=>2®e	L±² 
³p0/Z
2. .p0/ ´µ
R1SO	¶ ·=4.p¸=4LO€5¹z 1 ¹4.p=4LWXY4g“œ`L¤o=§f
R2S†WXY42 Xª o4(8oS¥¦XYf
R3S*+
𝑎
𝑏=
𝑎
𝑏(𝑎≥0，𝑏> 0)LºWXY42®e=>f
R4S=>%?.L.…=>2 	f
R5S»=./L¼?¢0/Z
知识点9  二次根式的加减运算
 ½¾ F-g<¯_™LIJ¿Q¶´µ¥¦
知识点10  二次根式的混合运算
1. 67
­¼ À<67ÁÂI²L¤7ŽYÃXYL¨7Ž«L0ˆ7½¾L%Ä	 ¤7Ä	ŗ Z
2. F4672 67A#_²Æ+z 67Z
3. À<67 ÇÈH¡¢0/8¼ PZ
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【培优篇】
【题型1  二次根式的概念及其有意义的条件】
n 1R24-25 ÉÊ˶·ÌÍÎÏ·ÐÑÒÓSÔÕ 3𝑥−6 +
6−3𝑥+𝑦= 2025L 2025𝑥𝑦 5pRS
AZ2025 3
BZ2025 2
CZ2024
DZ2025
Ö×B
=ؤÙ%&' ()1Ú𝑥 5L¨ÛÜ1Ú𝑦 5L0ˆÝ7 2025𝑥𝑦Z|Þßàá
â%&' ()LãÒäå2WXY4ghi4g§ æZ
秧
è&©
3𝑥−6 ≥0
6−3𝑥≥0
 
L
§3𝑥−6 ≥0©𝑥≥2f
§6−3𝑥≥0©𝑥≤2Z
∴𝑥= 2Z
O𝑥= 2ÛÜ 3𝑥−6 +
6−3𝑥+𝑦= 2025©
3 × 2−6 +
6−3 × 2 +𝑦= 2025L
§©𝑦= 2025Z
∴2025𝑥𝑦=
2025 × 2 × 2025 =
20252 × 2 = 2025 2Z
éê
BZ
ë 1-1R24-25 ÉÊ˶·’ÍÍì·ÐÑÒÓS¶íž2LIEg gRîîS
AZ −2
BZ
3 3
CZ 5
DZ 𝑎+ 1
Ö×C
=Ø|àáâ E'LPQ 𝑎(𝑎≥0) TULÙE'ïœ=ØvjZ
秧
AZ∵−2 < 0L∴−2®gLéŸêœ®ð<&f    
BZ∵
3 3 ñ4g 3L∴
3 3®gLéŸêœ®ð<&f    
CZ∵5 > 0L∴5gLéŸêœð<&f
DZ•𝑎+ 1 < 0v𝑎< −1`L 𝑎+ 1®gLéŸêœ®ð<&f    
éê
CZ
ë 1-2R24-25 ÉÊ˶·òóôõ·XàöSÔÕ
𝑦=
𝑥−4+ 4−𝑥
3
−2L3𝑥+ 4𝑦=      
Z
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Ö×4
=ØŸÞßàáâ%&' ()L÷ø§®yùg§æZ
*+%&' ()©Ú®yù1Ú xLy 5LúˆÛÜÝ7Z
秧
∵𝑦=
𝑥−4+ 4−𝑥
3
−2L
∴
𝑥−4 ≥0
4−𝑥≥0
 
L
§©
𝑥= 4L
é𝑦= −2L
3𝑥+ 4𝑦= 3 × 4 + 4 × (−2) = 4Z
éÖ×p
4Z
ë 1-3R24-25 ÉÊ˶·òóôõ·XàöSŒ𝑥Ã𝑦pF4L°ab2 4−𝑥+ |𝑦| = 𝑥−4L 𝑥−𝑦 
7wxYp      
Z
Ö× 2
=Ø|àá%&' ()Ãhi4 ,-Ã17wxYL¤Ù7wxY€5 
hi,LÇ<%&' ()1© xÃy 5L¥û1© 𝑥−𝑦vj1§Z
秧
∵2 4−𝑥+ |𝑦| = 𝑥−4L2 4−𝑥≥0L|𝑦| ≥0L
∴4−𝑥≥0°𝑥−4 ≥0L
∴4−𝑥= 𝑥−4 = 0L§©𝑥= 4L
†𝑥= 4ÛÜ2 4−𝑥+ |𝑦| = 𝑥−42L©𝑦= 0L
∴𝑥−𝑦=
4−0 =
4 = 2L
∴𝑥−𝑦 7wxYp 2Z
éÖ×p
 2Z
【题型2  利用二次根式的性质化简】
n 2R24-25 ÉÊ˶·üýþÿ·ÐÑÒÓSO=𝑎−
1
𝑎L	‡ l> a ‰¥	‹ ÇÈgR   S
AZ 𝑎
BZ −𝑎
CZ−𝑎
DZ−−𝑎
Ö×D
=ØÞßàáâ &'Z§ æg ÷ø O	‡ Û484l‰Š	‹ĀL{
gXY ā67LĂ	‡‰Š	‹ßxYL¯°‰Š	‹­ă„	‹ TgŽ ˜Z&
	‡ 4l8Tgi4`LÛĄ¼dTgi5LßOi	ąŠ	‡¨xYˆ‰Š	‹Z
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QÈ	‡ 4l8Tgi4`LÛĄ¼dTgi5LßOi	ąŠ	‡¨xYˆ‰Š	‹Lú
ˆ./vjZ
秧
è &'jÕ𝑎< 0L
∴𝑎−
1
𝑎= −−
1
𝑎× (−𝑎)2 = −−𝑎Lé D ÷øZ
éê
DZ
ë 2-1R24-25 ĆÊËć·ĈĉÌ·ÐÑÒÓSŒ𝑥𝑦> 0L −𝑥2𝑦./pR  S
AZ𝑥−𝑦
BZ−𝑥𝑦
CZ−𝑥𝑦
DZ−𝑥−𝑦
Ö×D
=Ø|àáâ ./LãÒäå ,-g§æZ¤ĊċÚ𝑥< 0L𝑦< 0L¨*
+ ,-./vj©Z
秧
∵𝑥𝑦> 0L
∴𝑥,𝑦_	L°Č®p 0L
č∵Ď −𝑥2𝑦2L−𝑥2𝑦gWXY4L
∴−𝑥2𝑦> 0L
∴𝑦< 0L
∴𝑥< 0L
∴−𝑥2𝑦= |𝑥| −𝑦= −𝑥−𝑦L
éê
DZ
ë 2-2Œ𝑚< 0LOT𝑚𝑥	‡— o‰Š	ň THgR S
AZ−𝑚2𝑥
BZ −𝑚2𝑥
CZ (−𝑚)2𝑥
DZ−−𝑚𝑥
Ö×A
=ØÙ m<0L¼ďgi4L m ‰Š	‹ĀˆL‡—%i	L¥û©ÚÖ×.
秧
∵m<0,
∴OT𝑚𝑥	‡— o‰Š	ň Tp
−𝑚2𝑥.
éê A.
ĐđŸÞßàáâ ./15LÙ¼ď ð	©Úg§ æ.
ë 2-3Œ 1575𝑛g¼4L¡ÚId÷¼4𝑛 j 5      
Z
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Ö×7RÖ×®ĒIS
=ØßēĔ1575𝑛 XxYp¼4L†1575o4=§L¥ûvj1§Z
秧
∵1575 = 9 × 25 × 7 = 32 × 52 × 7L
Œ 1575𝑛g¼4L𝑛 5jkg7
éÖ×p
7RÖ×®ĒISZ
ĐđŸÞßàáâ ,-L÷ø./©Úg§æZ
【题型3  最简二次根式】
n 3R24-25 ÉÊËć·ĕĖėó·ÐÑÒÓSÔÕ 𝑥2 + 1g0/L°­ 5jk<¯L𝑥 5
pR    S
AZ2
BZ−4
CZ4
DZ ± 2
Ö×D
=Ø|àáâ_™L0/L*+xY§YĘLãÒäåÕęĐ H+g§ 
æZ
è 𝑥2 + 1g0/°­ 5jk<¯L©Ú𝑥2 +1 = 5Lúˆ*+xY§YĘvjZ
秧
∵𝑥2 + 1g0/°­ 5jk<¯L
∴𝑥2 +1 = 5L§©
𝑥=± 2L
éê
DZ
ë 3-1R24-25 ÉÊËć·ćĚěĜ·ĝĞS¶í2Lg0/ gR   S
AZ 18
BZ 63𝑎
CZ 𝑎2−2𝑎+ 4
DZ
𝑎
3
Ö×C
=Ø|àáâ0/ E'LĊEIdg®g0/ YLğgïdĠá
0/ cd()gġ_`abL_`ab ğg0/Lġğ®gZ
ab¶ícd() LUV0/
R1SWXY4 o4g¼4Log¼f
R2SWXY42®e X©ªY o48oZ
jkŸ„ĊċĢdêœg÷ø Z
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秧
Aà 18=3 2L®g0/Lé|Ꜯð<&f
Bà 63𝑎= 3 7𝑎L®g0/Lé|Ꜯð<&f
Cà 𝑎2−2𝑎+ 4g0/Lé|êœð<&f
DÃ
𝑎
3 =
3𝑎
3 L®g0/Lé|Ꜯð<&f
éê
CZ
ë 3-2†T 35−𝑎Ra p÷¼4S.p0/ˆLjk­ 8<¯Z¡ÚIdð<() a 
5      
Z
Ö×3RÖ×®ĒIS
=Ø|àáâ_™L0/LãÒäå_™ E'g§ æZ
秧
∵8 = 2 2L
∴35−𝑎jkp 2L2 2L3 2L4 2L
∴35−𝑎= 2835−𝑎= 8835−𝑎= 18835−𝑎= 32L
§©
𝑎= 338𝑎= 278𝑎= 178𝑎= 3L
éÖ×p
3Z
ë 3-3ÔÕ𝐴= 2 2𝑥+ 1L𝐵= 3 𝑥+ 3L𝐶=
10𝑥+ 3𝑦L$2 ALB p0/L°
𝐴+ 𝐵= 𝐶L2𝑦−𝑥 5p      
Z
Ö×68
=ØÙ&©Ú2𝑥+ 1 = 𝑥+ 3L1Ú𝑥= 2L¥û©Ú10𝑥+ 3𝑦= (5 5)
2
= 125L1Ú𝑦= 35L¨
ÛÜ15vjZ
ç§∵ALB p0/L°𝐴+ 𝐵= 𝐶L
∴2𝑥+ 1 = 𝑥+ 3L
§©𝑥= 2L
∴𝐴= 2 5L𝐵= 3 5L𝐴+ 𝐵= 5 5 = 𝐶L
∴10𝑥+ 3𝑦= (5 5)
2
= 125L
§©𝑦= 35L 
∴2𝑦−𝑥= 2 × 35−2 = 68Z
éÖ×p
68Z
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Đđ|àá0/LÙ0/ E'©Úx = 2g§ æZ
【题型4  求二次根式中的参数值】
n 4Œ(2−7)
2
= 𝑎−𝑏7L$2𝑎L𝑏Čp¼4L𝑏−𝑎 5p      
Z
Ö×−7
=Ø|àᎏ67­Û415Lģ#ĤĥxYĦ#”yL*+ĤĥxY
Ħ\XˆLè”y ()©Š𝑎L𝑏LÛÜĄħ1§vj©ŠÖ×LãÒäå67
g§Ĩĩ æZ
秧
 ∵(2−7)
2
= 22−2 × 2 ×
7 + ( 7)
2
= 11−4 7 = 𝑎−𝑏7L
∴𝑎= 11,𝑏= 4L
∴𝑏−𝑎= 4−11 = −7L
éÖ×p
−7Z
ë 4-1 2𝑏+ 1­ 𝑎−1  p 0L𝑎+ 𝑏 5p     Z
Ö×
1
2/0.5
=Ø|àáâ hi,L1¼ 5fj© 2𝑏+ 1 +
𝑎−1 = 0Lè hi,©
𝑎−1 = 0L2𝑏+ 1 = 0L1Ú𝑎𝑏LÛ5vj1§f?§ hi, 𝑎≥0R𝑎≥0Sg§ æZ
秧
è&©
2𝑏+ 1 +
𝑎−1 = 0L
∴ 𝑎−1 = 0L2𝑏+ 1 = 0L
§©
𝑎= 1L𝑏= −
1
2L
∴ 𝑎+ 𝑏
= 1−1
2
=
1
2f
éÖ×
1
2Z
ë 4-2R2025 ĆÊ˶·Īī·ĬÒÓSŒ
𝑚
2 = 2 2L¼4𝑚 5p           
Z
Ö×16
=Ø|àá 67LcĭxY1Ú m 5vj§Z
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秧
∵
𝑚
2 = 2 2L
∴
𝑚
2 = 8L
§©𝑚= 16L
éÖ×p
16Z
ë 4-3ÔÕ 10−𝑛g¼4L÷¼4𝑛q%j 5Į%      
dL{N p      
Z
Ö×     
4     
26
=Ø|àá%&' ()Ã ,-LÙ WXY4ghi4Ç<
 ,-©Š÷¼4 n 5L¥û1§vjZ
秧
Ď 10−𝑛2L10−𝑛≥0L𝑛≤10L
∵10−𝑛g¼4L° n p÷¼4Lč 0 = 0L 1 = 1L 4 = 2L 9 = 3L 
∴n 5p 10 8 9 8 6 8 1LĮ 4 dL
{N p10 + 9 + 6 + 1 = 26L
éÖ×p
4L26Z
【题型5  二次根式的运算或化简求值】
n 5ÔÕ𝑥=
2 +1L𝑦=
2−1L1¶íž 5Z
(1)𝑥2 + 𝑦2f
(2)
𝑥
𝑦+
𝑦
𝑥Z
Ö×(1)6
(2)2 2
=Ø|àá ./15LãÒäå À<67g§ æZ
R1S*+𝑎2 + 𝑏2 = (𝑎+ 𝑏)2−2𝑎𝑏LÛÜ𝑥=
2 +1L𝑦=
2−1Lvj15f
R2S¤©Ú(
𝑥
𝑦+
𝑦
𝑥)
2
=
𝑥
𝑦+2
𝑥
𝑦⋅
𝑦
𝑥+
𝑦
𝑥=
𝑥
𝑦+
𝑦
𝑥+2 =
𝑥2+𝑦2
𝑥𝑦
+2LÇ<R1SÛÜ5vj1©Ö×Z
ç§R1S§
Ù𝑥=
2 +1,𝑦=
2−1L%
𝑥+ 𝑦= ( 2 +1) + ( 2−1) = 2 2L
𝑥𝑦= ( 2 +1)( 2−1) = 2−1 = 1L
ÛÜ𝑥2 + 𝑦2©Š
𝑥2 + 𝑦2 = (𝑥+ 𝑦)2−2𝑥𝑦= (2 2)
2
−2 × 1 = 8−2 = 6f
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R2S(
𝑥
𝑦+
𝑦
𝑥)
2
=
𝑥
𝑦+2
𝑥
𝑦⋅
𝑦
𝑥+
𝑦
𝑥=
𝑥
𝑦+
𝑦
𝑥+2 =
𝑥2+𝑦2
𝑥𝑦
+2L
op𝑥2 + 𝑦2 = 6L𝑥𝑦= ( 2 +1)( 2−1) = 2−1 = 1L
qk(
𝑥
𝑦+
𝑦
𝑥)
2
=
𝑥2+𝑦2
𝑥𝑦
+2 =
6
1 +2 = 8L
qk
𝑥
𝑦+
𝑦
𝑥=
8 = 2 2Z
ë 5-1Ý7
(1) 3 +3
1
3−12f
(2) 45 ÷
27 ×
3
5f
(3) 18 ÷
2−32 +
3 ×
6f
(4)(3 +
3)(3−3) + (1 +
7)
2
Z
Ö×(1)0
(2)1
(3)3−2
(4)14 + 2 7
=ØŸàáâ À<67LãÒäå 67gæZ
R1SO./ˆ¥¦½¾67vjf
R2S¤O«į.pŽL¨¥¦Ý7vjf
R3S¤Ý7Ž«L¨Ý7½¾vjf
R4S*+xYĦĤĥxY¥¦Ý7vjZ
ç§R1S§
 3 +3
1
3−12
=
3 +
3−2 3
= 0
R2S 45 ÷
27 ×
3
5
=
45
27 ×
3
5
=
45
27 × 3
5
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
=
1
= 1
R3S 18 ÷
2−32 +
3 ×
6
=
9−4 2 +
18
= 3−4 2 + 3 2
= 3−2
R4S(3 +
3)(3−3) + (1 +
7)
2
= 32−( 3)2 + (1 + 2 7 + 7)
= 9−3 + 1 + 2 7 + 7
= 14 + 2 7
ë 5-2R24-25 ÉÊËć·ĈħÏ·ÐÑÒÓSÔÕ𝑥= 2−3L1Û4(7 +
3)𝑥2 + (2 +
3)𝑥+
3
 5Z
Ö×38−20 3
=Ø|àá gÔÕ()L1§Û4 5L¤1§𝑥2 = (2−3)
2
= 7−4 3L¨ÛÜ¥¦Ý7v
jZ
秧
∵𝑥= 2−3L
∴𝑥2 = (2−3)
2
= 7−4 3L
∴(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 3Z
ë 5-3ÔÕ 𝑥+ 𝑦= −5L𝑥𝑦= 4L1𝑥
𝑦
𝑥+𝑦
𝑥
𝑦 5Z
Ö×−4
=ؤÙ𝑥+ 𝑦= −5L𝑥𝑦= 4LjĊċ𝑥< 0L𝑦< 0L¨†ă./Lúˆ†ÔÕ()¼ďÛÜ15
vjZ
|Þßàáâ./# ,-LãÒäå 𝑎2 = |𝑎| =
𝑎(𝑎> 0)
0(𝑎= 0)
−𝑎(𝑎< 0)
 
g§ æZ
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
ç§ ∵𝑥+ 𝑦= −5 < 0L𝑥𝑦= 4 > 0L
∴𝑥< 0L𝑦< 0L
∴ă=𝑥
𝑥𝑦
𝑥2 +𝑦
𝑥𝑦
𝑦2 = −𝑥⋅
𝑥𝑦
𝑥−𝑦⋅
𝑥𝑦
𝑦= −2 𝑥𝑦 .
∵𝑥𝑦= 4Z
ă = −2 ×
4 = −2 × 2 = −4Z
【拔尖篇】
【题型6  复合二次根式的化简】
n 6İı¶íž
5 + 2 6 = (2 + 3) + 2 2 × 3 = ( 2)
2
+ ( 3)
2
+2 2 ×
3 = ( 2 +
3)
2
L
8 + 2 7 = (1 + 7) + 2 1 × 7 = 12 + ( 7)
2
+2 × 1 ×
7 = (1 +
7)
2
L……ZIJ6+kć Y./
7 + 2 10 =        
Z
Ö× 5 +
2/ 2 +
5
=Ø|àáâ;< ./LĤĥxY H+f¿ij2Ĵĵ Y¥¦./vjZ
秧
 7 + 2 10 =
5 + 2 + 2 10
=
( 5)2 + ( 2)2 + 2 ×
5 ×
2
=
( 5 +
2)2
=
5 +
2f
éÖ×p
 5 +
2Z
ë 6-1R24-25 ĆÊ˶·ÌÍĶÏ·XàöS|2−3| +
4−2 3 = R    S
AZ2 3−3
BZ2 3−1
CZ3
DZ1
Ö×D
=ØŸÞßàáâ À<67L*+€5 &'ŽÇ< ,-¥¦.
/Z
秧
|2−3| +
4−2 3
= |2−3| +
( 3)2−2 3 + 1
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
= 2−3 +
( 3−1)2
= 2−3 +
3−1
= 1L
éê
DZ
ë 6-2ķ𝑎p−13 +
5−3−5 ¹4Ā=L𝑎= RîîS
AZ
7
5
BZ
2
2
CZ 2−1
DZ 3−1
Ö×C
=Ø|àáĸ?4 Ĺ5Z*+ĺĻ¤†ăëPLúˆ/.Ý7ÇÈL0ˆĹÝÚ¹4Ā= 
5LúˆĂêœ2¥¦áļL0Ľľ vpÖ×Z
秧
|Ù()L𝑎p−13 +
5−3−5 ¹4Ā=LoŸ𝑎< 1LoŸjkĿ« AÃD êœZ
ķ𝑥=
5−3−5L
∵3−5 =
6−2 5
2
=
( 5−1)
2
2
L
∴3−5 =
( 5−1)
2
2
=
5−1
2 =
10−2
2
L
∴𝑥=
5−
10−2
2
=
2 5−10+ 2
2
L
∵5 ≈2.24L 10 ≈3.16L 2 ≈1.41L
∴𝑥=
2 5−10+ 2
2
= 2 × 2.24−3.16 + 1.41
2
= 4.48−3.16 + 1.41
2
2.73
2 = 1.365L
∵−13 +
5−3−5 = −13 + 𝑥= −11.635L
∵−11.635 ¼4Ā=g−12L
∴¹4Ā=p−11.635−(−12) = 0.365L
êœ B g
2
2 ≈0.707Lêœ C g 2−1 ≈0.41L
ŀ%êœ C 0ĽľÖ×Z
éê
CZ
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
ë 6-3R24-25 ÉÊËć·ŁłŃÏ·ĝĞS./ 23−6 10 + 4 3−2 2 ÇÈgR   S
AZ3 +
2
BZ3−2 2
CZ3 + 2 2
DZ3−2
Ö×D
=Ø|àá 67LÙ ,-/Ç<*+ĤĥxYÝ7vj§Z
秧
ă =
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
= 3−2L
éê
DZ
【题型7  利用分母有理化求值】
n 7R24-25 ÉÊËć·Ĉ¢ń·ĝ2SŅņ¶—Ý7ŇĘ
1
2+1 =
1×( 2−1)
( 2+1)( 2−1) =
2−1f
1
3+ 2 =
1×( 3−2)
( 3+ 2)( 3−2) =
3−2f
1
5+2 =
1×( 5−2)
( 5+2)( 5−2) =
5−2Z
ö1
(1)
1
7+ 6 5p Z
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
(2)1
1
1+ 2 +
1
2+ 3 +
1
3+ 4 +⋯+
1
2022+ 2023 +
1
2023+ 2024 5Z
(3)Œ𝑎=
1
5−2L1𝑎3−4𝑎2 +𝑎+ 4 5Z
Ö×(1) 7−6
(2)2 506−1
(3)2 5 +8
=ØŸàáâ=>%?.à ./15Lňʼn=>%?. Yg§| æZ
R1SăÙŅņŊŋ2 YëPvj©ŠÇÈf
R2SăžœëPˆLŌō<¯vj©ŠÇÈf
R3S¤./ aLúˆÛÜq1TÝ7vjZ
ç§R1S
1
7+ 6 =
7−6
( 7+ 6)( 7−6) =
7−6f
R2Să =
2−1 +
3−2 +
4−3 +⋯+
2023−2022 +
2024−2023
=
2024−1
= 2 506−1f
R3S∵𝑎=
1
5−2 =
5+2
( 5−2)( 5+2) =
5 +2L
∴𝑎−2 =
5L
∴(𝑎−2)2 = 5L
∴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 +8Z
ë 7-1ŎŇƕ67L†=>2 .p%? ŇĘL³p=>%?.Z
Q
1
2 =
2
2 L
1
3−2 =
1×( 3+ 2)
( 3−2)R 3+ 2S
=
3 +
2±ŏ67ń³p=>%?.Z
(1)†¶í=>?.
2−3
12 = ___________L
𝑥−𝑦
𝑥−𝑦= ___________
(2)ŐÃőcŒ./
𝑥−𝑦
𝑥+ 𝑦`L¡Úcœ®_ §ÖŇĘZ
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
Ő
𝑥−𝑦
𝑥+ 𝑦=
(𝑥−𝑦)
𝑥−𝑦
𝑥+ 𝑦
𝑥−𝑦=
(𝑥−𝑦)
𝑥−𝑦
𝑥−𝑦
=
𝑥−𝑦f
ő
𝑥−𝑦
𝑥+ 𝑦=
( 𝑥)
2−( 𝑦)2
𝑥+ 𝑦
=
𝑥+ 𝑦
𝑥−𝑦
𝑥+ 𝑦
=
𝑥−𝑦
IJŔŕŖŅņŐÃőcŒ §ŇĘL€ŐÃőcŒ §ÖÚŗĊRîîS
AZՀőŘ    BZŐŘő€    CZŐÃőĥ€    DZŐÃőĥŘ
(3)ÔÕ%?4 aÃb ab
𝑎
2+1 +
𝑏
2 = −1 + 2 2L1 aÃb 5Z
Ö×(1)
2 3−3
6
L 𝑥+
𝑦
(2)C
(3)𝑎= 1L𝑏= 2
=Ø|Þßàáâ=>%?.L§ ægļř%?.oZ
R1Sَk%?.o8ÙxYĦo=§./Ý7vjf
R2SÙR1S2Y¥¦Ċċvjf
R3SÙYIL¥¦=>%?.Ý7©Ú 𝑎+
1
2 𝑏
2−𝑎= −1 + 2 2LÙ𝑎,𝑏p%?4L¥ûvj1©
𝑎,𝑏 5Lvj1§Z
ç§R1S§
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
éÖ×p
2 3−3
6
L 𝑥+
𝑦f
R2SÙR1S2 Y¥¦Ý7jÕLŐÃőń€
éê
CZ
R3S§
 ∵
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
∵𝑎L𝑏 g%?4
∴𝑎+ 1
2 𝑏= 2L−𝑎= −1
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
∴𝑎= 1
∴1 + 1
2 𝑏= 2
∴𝑏= 2Z
ë 7-2R24-25 ÉÊËć·ĪīŚE·ĝ2SŅņŊŋ
Ď¥¦./`LMN%`śŜćQ
3
5L
2
3L
2
3+1±²I™ TL$FMNŝjk†$¥I´.
/
3
5 =
3× 5
5× 5 =
3
5 5f
2
3 =
2×3
3×3 =
6
3 f
2
3+1 =
2×( 3−1)
( 3+1)( 3−1) =
2( 3−1)
( 3)
2−12 =
3−1Zkć±œ./ ´µU
V=>%?.Z
§Ĩĩ
(1)Şijć— §ŇĘL./
1
7−6 = __________Z
(2)Ý7
1
2+1 +
1
3+ 2 +
1
4+ 3 + ⋯+
1
2025+ 2024 × ( 2025 + 1)Z
(3)ÔÕ𝑎=
1
3+ 2L𝑏=
1
3−2L1𝑎2 + 𝑏2 5Z
Ö×(1) 7 +
6
(2)2024
(3)10
=Ø|àá=>%?4L À<67L./15
R1SÙ=>%?. Y¥¦1§vjf
R2S¤¥¦=>%?.L¨¥¦Ý7vjf
R3S¤¥¦=>%?.1Ú𝑎,𝑏 5L¥û1Ú𝑎+ 𝑏,𝑎𝑏 5L*+ĤĥxYëP15vjZ
ç§R1S§
1
7−6 =
7+ 6
( 7−6)( 7+ 6) =
7+ 6
7−6 =
7 +
6f
éÖ×p
 7 +
6f
R2Să = (−1 +
2−2 +
3 + ⋯−2024 +
2025) × ( 2025 + 1)
= ( 2025−1) × ( 2025 + 1)
= 2025−1 = 2024f
R3S∵𝑎=
1
3+ 2 =
3−2
( 3+ 2)( 3−2) =
3−2L
𝑏=
1
3−2 =
3+ 2
( 3−2)( 3+ 2) =
3 +
2L
∴𝑎+ 𝑏= 2 3L𝑎𝑏= 1L
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
∴𝑎2 + 𝑏2 = (𝑎+ 𝑏)2−2𝑎𝑏= (2 3)
2
−2 = 10Z
ë 7-3R24-25 ÉÊËć·uş·ĝ2SnQ
2+1
2−1 =
( 2+1)( 2+1)
( 2−1)( 2+1) = 2 2 +3ZŠ±²ŎŇ=TÃ=>_
ŽIdTO=>2 	.…8šO	2 =>.…LUV=>%?.Z%¶íÇŢ
①Œ a g 3 ¹4Ā=L
3
𝑎 5p3 3 +6f
②šţŤ¹
 2024−2023 <
2023−2022f
③ëP
1
3+ 2+1 =
2+ 2−6
4
f
④Ý7
2
3+ 3 +
2
5 3+3 5 +
2
7 5+5 7 +⋯+
2
99 97+97 99 = 1−
11
33 f
⑤ÔÕ𝑎=
𝑚+1−𝑚
𝑚+1+ 𝑚L𝑏=
𝑚+1+ 𝑚
𝑚+1−𝑚L°𝑎2 +1926𝑎𝑏+ 𝑏2 = 2024Lq%j  ¼4 m p−1Z
kćÇŢ÷ø gR    S
AZ①③④⑤
BZ②③④
CZ②③⑤
DZ②③④⑤
Ö×B
=Ø|àá*+= ť|,-ÃxYĦ¥¦=>%?.L§Ĩ ./ÚţŤ¹
67 ĩLãÒäåÕęĐg§ æZ
①ĹÝĸ?4 ¼4Ā=L1Ú¹4Ā=𝑎=
2−1L¥û=>%?.¥¦./f
②ŎŇ=>%?.Lšţcd Ŧ4Ť¹Lvj§Öf
③¤=T=>_`Žk( 3−2−1)L¾ŧ=> d4ˆ¨%?.=>vjf
④ŎŇ=>%?.ļŠ2ĸ?1 67@ALĂû./1Ú5f
⑤𝑎­ b jk*+=>%?../Lj©Ú𝑎𝑏= 1LúˆİıYĘŨĐL1© m 5Z
秧
①∵a g 3 ¹4Ā=L
∴𝑎=
3−1L
∴
3
𝑎=
3
3−1 =
3( 3+1)
( 3−1)( 3+1) =
3( 3+1)
2
=
3 3+3
2
Lé①Řũf
②∵
1
2024−2023 =
( 2024−2023)( 2024+ 2023)
( 2024−2023)
=
2024 +
2023L
1
2023−2022 =
( 2023−2022)( 2023+ 2022)
( 2023−2022)
=
2023 +
2022L
č∵2024 +
2023 >
2023 +
2022L
∴
1
2024−2023 >
1
2023−2022L
∴2024−2023 <
2023−2022Lé②÷øf
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
③
1
3+ 2+1
=
( 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
Lé③÷øf
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 Lé④÷øf
⑤∵𝑎=
𝑚+1−𝑚
𝑚+1+ 𝑚=
( 𝑚+1−𝑚)( 𝑚+1−𝑚)
( 𝑚+1+ 𝑚)( 𝑚+1−𝑚) = ( 𝑚+ 1−𝑚)
2
L
∴𝑎> 0L
∵𝑏=
1
𝑎L
∴𝑎𝑏= 1L𝑏=
𝑚+1+ 𝑚
𝑚+1−𝑚= ( 𝑚+ 1 +
𝑚)
2
L
∴𝑏> 0L
教辅资源，关注公众号★全科AA+
                                                                           
                                                                       
∴𝑎+ 𝑏> 0L
∵𝑎2 +1926𝑎𝑏+ 𝑏2 = 2024L
∴𝑎2 +1926 + 𝑏2 = 2024L
𝑎2 + 𝑏2 = 98L
𝑎2 + 𝑏2 +2𝑎𝑏= 98 + 2𝑎𝑏L
(𝑎+ 𝑏)2 = 98 + 2L
(𝑎+ 𝑏)2 = 100L
∵𝑎+ 𝑏> 0L
∴𝑎+ 𝑏= 10L
v( 𝑚+ 1−𝑚)
2
+ ( 𝑚+ 1 +
𝑚)
2
= 10L
4𝑚+ 2 = 10L
§©𝑚= 2Lé⑤ŘũZ
Ūćqū
②③④÷øL
éê
BZ
【题型8  二次根式的规律探究】
n 8./
1
8+ 11 +
1
11+ 14 +
1
14+ 17 +
1
17+ 20 +
1
20+ 23 +
1
23+ 26 +
1
26+ 29 +
1
29+ 32ÇÈgR   S
AZ1
BZ
2 2
3
CZ2 2
DZ4 2
Ö×B
=Ø|àáâ=>%?.# ½¾67LãÒäå67g§ æZ¤†Ŭd=
¥¦=>%?.L¨Ý7½¾vj©ÚÖ×Z
秧
1
8+ 11 =
11−8
( 11+ 8)( 11−8) =
11−8
3
L
_?j©
1
11+ 14 =
14−11
3
⋅⋅⋅
1
29+ 32 =
32−29
3
L
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
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= 2 2
3
éê BZ
ë 8-1R24-25 ĆÊ˶·ÌÍŭŮ·ĝĞSİı¶íy
①3 + 2 2 = ( 2 + 1)
2
f②5 + 2 6 =
( 3 +
2)
2
f③7 + 2 12 = (2 +
3)
2
f…fIJÙkć@AL¡Úů 9 dy      
Z
Ö×19 + 2 90 = ( 10 + 3)
2
=Ø|àáâ4l™@ABŰL ,-LÙÔÕyűŲIJ@Ag§æZİıy
j©ů𝑛dyp(2𝑛+ 1) +2 𝑛(𝑛+ 1) = ( 𝑛+ 1 +
𝑛)
2
Lvj1§Z
秧
è&jÕLů 1 dy3 + 2 2 = ( 2 + 1)
2
Lv(2 × 1 + 1) +2 1 × 2 = ( 2 +
1)
2
f
ů 2 dy5 + 2 6 = ( 3 +
2)
2
Lv(2 × 2 + 1) +2 2 × 3 = ( 3 +
2)
2
f
ů 3 dy7 + 2 12 = (2 +
3)
2
Lv(2 × 3 + 1) +2 3 × 4 = ( 4 +
3)
2
f
……
İıűŲLů𝑛dyp(2𝑛+ 1) +2 𝑛(𝑛+ 1) = ( 𝑛+ 1 +
𝑛)
2
L
ů 9 dyp19 + 2 90 = ( 10 + 3)
2
L
éÖ×p
19 + 2 90 = ( 10 + 3)
2
Z
ë 8-2ŎŇ“èŨųŠIJ” YBC¶— 67@A
Ũn 1
 1 +
1
3 =
3+1
3 =
4 ×
1
3 = 2
1
3fŨn 2
 2 +
1
4 =
8+1
4 =
9 ×
1
4 = 3
1
4fŨn 3
 3 +
1
5 =
15+1
5
=
16 ×
1
5 = 4
1
5……H+űŲ 6
7@A1 2024 +
1
2026 ×
4052 5R      S
AZ2024
BZ2025 2
CZ2023
DZ2023 2
Ö×B
=Ø|àáļ@ALÙ2Ũn©Š@A 𝑛+
1
𝑛+2 =
𝑛(𝑛+2)+1
𝑛+2
=
(𝑛+ 1)2 ×
1
𝑛+2 = (𝑛+ 1)
1
𝑛+2L
Û51§vj©ŠÖ×LİıÔÕŨnŨŴLřøļŠ@Ag§Ĩĩ æZ
秧
Ũn 1
 1 +
1
3 =
3+1
3 =
4 ×
1
3 = 2
1
3f
Ũn 2
 2 +
1
4 =
8+1
4 =
9 ×
1
4 = 3
1
4f
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Ũn 3
 3 +
1
5 =
15+1
5
=
16 ×
1
5 = 4
1
5f
……
∴kć@Ap
 𝑛+
1
𝑛+2 =
𝑛(𝑛+2)+1
𝑛+2
=
(𝑛+ 1)2 ×
1
𝑛+2 = (𝑛+ 1)
1
𝑛+2L
∴•𝑛= 2024`L 2024 +
1
2026 ×
4052 = 2025
1
2026 ×
4052 = 2025 2L
éê
BZ
ë 8-3R24-25 ÉÊ˶·ÌÍÎÏ·ĝ2S¤İı¶íyL¨ŵÖĩ
①1 +
1
12 +
1
22 = 1 +
1
1−
1
2f
②1 +
1
22 +
1
32 = 1 +
1
2−
1
3f
③1 +
1
32 +
1
42 = 1 +
1
3−
1
4f
…
IJŔ*+űŲ @ALÝ7
1 +
1
12 +
1
22 +
1 +
1
22 +
1
32 +
1 +
1
32 +
1
42 +⋯+
1 +
1
20222 +
1
20232−2024 =      
Z
Ö×−
2024
2023/−1
1
2023
=Ø①1 +
1
12 +
1
22 = 1 +
1
1−
1
2f②1 +
1
22 +
1
32 = 1 +
1
2−
1
3f③1 +
1
32 +
1
42 = 1 +
1
3−
1
4L©Š
1 +
1
20222 +
1
20232 = 1 +
1
2022−
1
2023LíÝ7vjZ
|àáâ2@ABŰLF4 Ý7LãÒäå@ABŰg§ æZ
秧
①1 +
1
12 +
1
22 = 1 +
1
1−
1
2f
②1 +
1
22 +
1
32 = 1 +
1
2−
1
3f
③1 +
1
32 +
1
42 = 1 +
1
3−
1
4L
é 1 +
1
20222 +
1
20232 = 1 +
1
2022−
1
2023L
é 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
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=
1 + 1 + 1 + … + 1
⏟
2022
+ 1−1
2 + 1
2 −1
3 + 1
3 −1
4 + … +
1
2022 −
1
2023 −2024
= 2023−
1
2023 −2024
= −
2024
2023L
éÖ×p
−
2024
2023Z
【题型9  二次根式的新定义】
n 9R24-25 ÉÊ˶·ŁłŶŷ·ÐÑÒÓSE'
MN†( 𝑎+
𝑏)­( 𝑎−𝑏)³pI€“€Ÿ”Zo
p
( 𝑎+
𝑏)( 𝑎−𝑏) = ( 𝑎)
2
−( 𝑏)
2
= 𝑎−𝑏Ljk%Ź …ź	Lqk%IŏĩjkŎŇŻż“€Ÿ”
„§ĨZnQ
ÔÕ 18−𝑥−11−𝑥= 1L1 18−𝑥+
11−𝑥 5Ljk±²§Ö
op( 18−𝑥−11−𝑥) × ( 18−𝑥+
11−𝑥) = ( 18−𝑥)
2
−( 11−𝑥)
2
= 18−𝑥−11 + 𝑥 = 7Lčop 18−𝑥−
11−𝑥= 1Lqk 18−𝑥+
11−𝑥= 7Z
R1SÔÕ
 20−𝑥+
4−𝑥= 8L𝑥 5g           
f
R2SÝ7
1
3 1+ 3 +
1
5 3+3 5 +
1
7 5+5 7 +⋯+
1
2023 2021+2021 2023 =           
Ö×     
−5     
1
2−
2023
4046
=ØŸàáâ ,-Ã À<67Ã%&' ()ÃxYĦk#=
>%?.LãÒäå 67ŽžëPg§ æZ
R1SŞijnLíÚYĘùL§YĘùvj©ŠÖ×f
R2S*+xYĦL€ă¥¦ëPˆLvj©ŠÖ×Z
秧
∵( 20−𝑥+
4−𝑥)( 20−𝑥−4−𝑥) = ( 20−𝑥)
2
−( 4−𝑥)
2
= 20−𝑥−4 + 𝑥= 16L
° 20−𝑥+
4−𝑥= 8L
∴20−𝑥−4−𝑥= 2f
∵
20−𝑥+
4−𝑥= 8
20−𝑥−4−𝑥= 2
 
∴2 4−𝑥= 6L
./ˆcĭ_`xY©
4−𝑥= 9L
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∴𝑥= −5L
ſĠƀ
𝑥= −5găYĘ §f
éÖ×p
−5Z
R2S
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 Z
éÖ×p
1
2−
2023
4046 Z
ë 9-1R24-25 ÉÊ˶·ÌÍÎÏ·ĝ2S€z~&®”y cdF4𝑎,𝑏LE'67※Q¶
•𝑎< 𝑏
`L𝑎※𝑏= 2 𝑎+
𝑏L•𝑎≥𝑏`L𝑎※𝑏= 2 𝑎−𝑏LnQ5※2 = 2 5−2L¿ćū@ELÝ7 3※2 −
8※12 ÇÈpR   S
AZ4 3−5 2
BZ−5 2
CZ4 3−3 2
DZ3 2
Ö×B
=Ø|àá gF4 67LÙqƁ T1Ú3※28※12 5L¨Ù ½¾Ý7Y
¥¦Ý7vjZ
秧
è&©L
3※2 = 2 3−2L
8※12 = 2 8 +
12 = 4 2 +2 3L
3※2 −8※12
= 2 3−2−(4 2 + 2 3)
= 2 3−2−4 2−2 3
= −5 2L
éê
BZ
ë 9-2R24-25 ÉÊ˶·ĈƂõ·ÐÑÒÓSE'
op( 𝑎+
𝑏)( 𝑎−𝑏) = ( 𝑎)
2
−( 𝑏)
2
= 𝑎−𝑏L
jk%Ź …ź	LMN³( 𝑎+
𝑏)­( 𝑎−𝑏)pI€“€Ÿ”ZŒ 18−𝑥−11−𝑥= 1L 18−𝑥+
11−𝑥=     Z
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Ö×7
=Ø|Þßàá 67LãÒäå Ž67#2qƁ67g§ æZƃ
Õ 18−𝑥+
11−𝑥­ 18−𝑥−11−𝑥gI€“€Ÿ”LjÙ 18−𝑥+
11−𝑥= ( 18−𝑥+
11−𝑥)
( 18−𝑥−11−𝑥)./Ý7vjZ
秧
ÙŊŋjÕL 18−𝑥+
11−𝑥­ 18−𝑥−11−𝑥gI€“€Ÿ”L
∵18−𝑥−11−𝑥= 1L
∴18−𝑥+
11−𝑥= ( 18−𝑥+
11−𝑥)( 18−𝑥−11−𝑥)
= ( 18−𝑥)2−( 11−𝑥)2
= 18−𝑥−(11−𝑥)
= 7
éÖ×p
7Z
ë 9-3€z~&F4 mLnLŒE'D67𝑚⊗𝑛=
𝑚−𝑛(𝑚≥𝑛),
𝑚+
𝑛(𝑚< 𝑛)
 
LƁÚƄdƅ
①18 ⊗2 = 2 2f
②
1
1⊗2 +
1
2⊗3 +
1
3⊗4 + ⋅⋅⋅+
1
99⊗100 = 100 ⊗1f
③(𝑎⊗𝑏) ⋅(𝑏⊗𝑎) = |𝑎−𝑏|Z
kćƅ2÷ø d4gR    S
AZ0 d
BZ1 d
CZ2 d
DZ3 d
Ö×D
=Ø*+DE'¥¦Ý7ïIĊċvjZ
秧
∵18 > 2L
∴18 ⊗2 =
18−2 = 3 2−2 = 2 2L
qk①÷øf
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
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qk②÷øf
•𝑎≥𝑏`L(𝑎⊗𝑏) ⋅(𝑏⊗𝑎) = ( 𝑎−𝑏)( 𝑏+
𝑎) = 𝑎−𝑏= |𝑎−𝑏|L
•𝑎< 𝑏`L(𝑎⊗𝑏) ⋅(𝑏⊗𝑎) = ( 𝑎+
𝑏)( 𝑏−𝑎) = 𝑏−𝑎= |𝑎−𝑏|L
qk③÷øf
é÷ø p①②③L% 3 dL
éê DZ
Đđ|àáDE'L À<67LäåDE' 67g§ æZ
【题型10  二次根式的实际应用】
n 10QƆL†IƇƈƉƊ”ĽjkƋ¢Idƌp 
8𝜋Lƍp 
2𝜋 ƎPLŒ†±Ƈƈ\XuDƉ
Ɗ”ĽƋ¢IdƏPL£Ə —g R      S
AZ12π
BZ18π
CZ24π
DZ36π
Ö×B
=Ø|àáâ À<67LƎP ,-LƏ —LÙ&©ÚƐƌL¥û1©Ə
 ƑƒLÙƏ —Lvj1§Z
秧
±Ƈƈ Ɛƌp2( 8π +
2π) = 6 2π
∴†±Ƈƈ\XuDƉƊ”ĽƋ¢IdƏPLƑƒp
6 2π
2π = 3 2
∴—pπ(3 2)
2
= 18π
éê
BZ
ë 10-1QƆLĎŤ÷YPƓƔ2ƕƖcd¹÷YPLÔÕcd¹÷YP —=ƒp𝑆1 = 18L𝑆2
= 12LuƗĀ=gId÷YPL$—p 2LƘƙĀ= —p      
Z
Ö×8 6−8/−8 + 8 6
=Ø|àáâ H+L¤7ÚƄd¹÷YP ĭƌL¨©ŠŤ÷YP ĭƌLŎҗ 
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Ý7©ÇŢZ
秧
 ∵Ƅd¹÷YP —=ƒp 18Ã12Ã2L
∴Ƅd¹÷YP ĭƌ=ƒp 18 = 3 2à 12 = 2 3à 2L
èƆÕ
Ť÷YP ĭƌp
3 2 +2 3−2 = 2 2 +2 3L
∴𝑆空白= (2 2 + 2 3)2−(18 + 12−2)
= 8 + 12 + 8 6−(18 + 12−2)
= 8 6−8Z
éÖ×p
8 6−8Z
ë 10-2R24-25 ÉÊËć·ÌĖƚ2·ĝ2SűƛƜŎƝéˆLƜƞƟıŎƠÙơƢˆƢƣƤŇ ƥ
ƦĹÝƢƧ¦ƨ ƩƪLq+ ſƀg𝑣= 16 𝑑𝑓L$2𝑣ĄƫƢƩR›Ƭ
km hSL𝑑ĄƫơƢˆƢ
ƣƤŇ ƥƦR›Ƭ
mSL𝑓ĄƫƭƮo4ZĎƯƜŎƝéưá2LƱ©𝑑= 20mL𝑓= 1.2LƲƝƳ
Ƣ ƢƩŤ»g“ŧƴR 6 ≈2.45LÇÈƵøŠ1km/hS
Ö×ƲƝƳƢ ƢƩŤ»g78km/h
=Ø|Þßàáâ H+L./yÕęĐL†𝑑= 20L𝑓= 1.2ÛÜ𝑣= 16 𝑑𝑓v
j1ÚƲƝƳƢ ƢƩŤ»g“ŧLãÒäå6+ ,-./15g§ĨŸ æZ
秧
∵𝑑= 20L𝑓= 1.2ÛÜ𝑣= 16 𝑑𝑓L
∴𝑣= 16 ×
20 × 1.2 = 16 × 2 6 = 32 6L
∵6 ≈2.45L
∴𝑣≈32 × 2.45 ≈78(km h)L
Ö
ƲƝƳƢ ƢƩŤ»g78km/hZ
ë 10-3R24-25 ÉÊËć·łĖóƶ·ĝĞSƷƘƸƹgIœ®ƺƻ Ƽƽ¦pLÙƾCLƹƿĂƦK
—p h ǀ Ʒǁǂèǃ¶LǃŠK— `DŽp tLſŇFƀLűŲ𝑡=
ℎ
5(®àDždžLJ Ljlj)Z
(1)1ƹďĂ40m ƷƘǃŠK— `DŽf
(2)ÔÕĂƷƘNJǃ ƹďq· Nj(›Ƭ
J) = 10 × ƹď-Nj(kg) × Ʒƪ(m)LInj-Njp0.2kg ǍǎſŇ
3sǃĎKćL±njǍǎ϶ǃŇĘ2q· Nj%“Ťƴ
(3)ĎR2S ÇÈ2LŔ ©ŠǏǐǑƫƴ(
ǒǓĸǔǕŒďŀǖß65J  Nj)
Ö×(1)2 2s
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(2)90J
(3)ǗǘƷƘƸƹ
=Ø|àáâÛ4 15L H+L÷ø?§&g§ æZ
R1SOℎ= 40mÛÜ𝑡=
ℎ
5Ý7vjf
R2S𝑡= 3sÛÜ𝑡=
ℎ
51© hL¥ûÝ7vjf
R3Sèz90J>65JL鶀Œďż¢ǓǙLH£ǘǚƷƘƸƹ
ç§R1S§
Oℎ= 40mÛÜ𝑡=
ℎ
5©
𝑡=
40
5 = 2 2sL
Ö
ƹďĂ40m ƷƘǃŠK— `DŽp2 2sf
R2S§
𝑡= 3sÛÜ𝑡=
ℎ
5©
ℎ
5 = 3L
§©
ℎ= 45mL
ĂƷƘNJǃ ƹďq· Njp10 × 0.2 × 45 = 90JL
Ö
±njǍǎ϶ǃŇĘ2q· Nj%90Jf
R3S§
∵90J>65JL
∴€ŒŻ¢ǓǙL
éǗǘƷƘƸƹZ
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