第十章数的开方(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(华东师大)

第十章  数的开方（举一反三讲义）全章题型归纳
 2024
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 1   !"  ........................................................................................................................5
 2  #$ !" %& ............................................................................................................6
 3  ' !" ( ........................................................................................................................9
 4  )*+,- ..........................................................................................................................................11
 5  ./*+0- ......................................................................................................................................13
12 .................................................................................................................................................................15
 6  -3 +45678 ..................................................................................................................15
 7  )*+9:; ..................................................................................................................................17
 8  <*=>?9*=> ......................................................................................................................19
 9  )*@*ABC,$ ..........................................................................................................................21
 10 )*,-+D$ ..................................................................................................................................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.
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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|= 𝒂E𝒂≥𝟎F
−𝒂E𝒂< 𝟎F
62=|6|=6
(−6)2=|−6|=6
𝒂𝟐
( 𝐚)𝟐 = 𝐚E𝐚≥𝟎F
( 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=aG 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.HI* a +" +,-JKL"GMN a JKOL*.
2.L?L"+PQ
L
L"
,-R	
±
3
OL*
:,
ST*
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I*
0 "#
+

#$+,-#
efaG"#


知识点6  无理数
.UVWX9*JK./*Y
./*+Z[\]^_`ab
E1FLLVc+*+dD e./*Gf 2G
3 5gh
E2FijkπlHmn^π+*,f 2πG
π
2gh
E3F_.UVWX9*\]op+*,f 0.101 001 000 1…EqI 1 rstuvHI 0)gYT./*+9
*=>w*.Uh./*+9*=>VWXh./*Vxyz{>*+\]Y
知识点7  实数的概念及分类
1. |}
^/*?./*~€)*Y
2. >
)*^qb>‚ƒ
E1F„…†>
)*‡>€^/*?./*Y
E2F„ˆ78>
)*‡>€ˆ)*!0!7)*Y
知识点8  实数与数轴的关系
ˆ<*
ˆ>*
7<*
7>*
)*
ˆ)*
7)*
ˆ^/*
ˆ./*
7^/*
7./*
0
)*
^/*
0
./*
ˆ^/*
7^/*
ˆ./*
7./*
^U9*ŠWX9*
.UVWX9*
ˆ<*
ˆ>*
7<*
7>*
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‹HI)*Œ‡_$*A+HIŽyzh‘G*A+‹HIŽŒyzHI)*Y)*@*A
+ŽHH’D.
知识点9  实数范围内的有关概念
“
8”
•–
d*
— a @ b ˜€d*G™𝑎+ 𝑏= 0
3+d*e−3
š*
— a @ b ˜€š*G™𝑎𝑏= 1
2 +š*e
1
2
S›)*+œ’(Œe67*G|𝑎| ≥0
|𝑎|
𝑎(𝑎> 0)
0(𝑎= 0)
−𝑎(𝑎< 0)
œ’(
˜€d*+qI*+œ’(dgG|𝑎| = |−𝑎|
| 2| = |−2| =
2
知识点10  实数的运算
ž)*Ÿ ¡GV¢‡_£¤¥!¦!§!¨E¨*V€©F!§,-Gª«‡_£¤L",-_l
67)*+L,-Y
^/*+,-8”?,-¬ž)*Ÿ ¡­®¯$G)*°C,-+±²@^/*°C,-+±²d³G
´§!LGµ§¨G¶·¥¦Y³¸,-„¹º»¼+±²£¤G^½	+´-½	¾¿+Y
知识点11  实数的大小比较
^/*9:;+Àž)*Ÿ ¡­®¯$Y
q)*+9Áf`
ˆ)*Œ 0G7)*Œ9 0Gˆ)*ÂHÃ7)*hqIˆ)*Gœ’(
+ˆ)*hqI7)*Gœ’(+7)*9hž*Ayz+qI)*G¼Ä+*Å:ºÄ+*
YÆÇG:;q)*+9È^f`À
E1Fɑ:;q)*++9G£ªÊ…)*+9ÁYf:; 10@ 3 +9G'Â( 10)
2 = 10
，32 = 9，10 > 9GË 10 > 3Y
E2F$0-+Àp./*+ÌÍ(·Gµ:;q*+9Y
E3FÎqIÏ 	+./*:;9ÐG‡D$f`ÑÒ
① 𝑎> 𝑏≥0⇔𝑎>
𝑏Y
② 𝑎> 𝑏⇔
3 𝑎>
3 𝑏Y
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【培优篇】
【题型1  直接求平方根、立方根】
– 1E24-25 ÓÔ¸`·ÕÖ×·ØÙF`ÚÛÀNÜÝ+eE    F
AY 5e 5 +-3 
BY0 + ?" Œe 0
CY 9+ e±3
DY−3e(−3)2+HI 
ÞßC
>àáâã @-3 +|}GäåH>àæçè+ˆÊ8Y
é%%
çè A
 5e 5 +-3 G
-3 e67*G« 5êë( 5)
2
= 5GË A ˆÊh
çè B
0 + ?" Œe 0G
 ?" +…†NG0 + ?" ì€ 0GË B ˆÊh
çè C
 9+ e ± 3G
9 = 3Gª 3 + D€ ±
3Gª6 ± 3G ± 3+e 9Gí 9+ °îÜÝGË C ÜÝh
çè D
−3e(−3)2+HI G
(−3)2 = 9G9 + € ± 3GïÆ−3e 9 +HI GË D ˆÊh
Ëç CY
ð] 1-1E24-25 ÓÔ¸`·ñòóô·õö÷øF−
8
27+" e   h 4+-3 e     Y
Þß     
−
2
3     
2
>àáâã" G-3 G%+ùeúû
füHIˆ*𝑥+gÂ𝑎GýþÿIˆ*𝑥
JK𝑎+-3 hfüHI*𝑦+"gÂ𝑏GýþÿI*𝑦JK𝑏+" YĀÆ%ޝ‡Y
é%%
−
8
27+" e−
2
3h
∵4 = 2G2+-3 e 2G
∴4+-3 e 2Y
ËÞ߀
−
2
3h 2Y
ð] 1-2E24-25 ÓÔ¸`·āĂ·ăØĄąFE1FĆy
a
0.000001
0.001
1
1000
1000000
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3 𝑎
E2F ĀćĈĉ+Ċ¬Ćċ
①
3 3 = 1.442G™
3 3000 = ______G
3 0.003 = ______h
②Čč
3 343 = 7G™
3 0.000343 = ______Y
ÞßE1F0.01G0.1G1 G10 G100E2F①14.42G0.1442G ②0.07
>àáĎďâãĐ" +8”GtĀOL*9*ŽđºŠđ¼Ēē 3 w’D+" +9*Ž
đºŠđ¼Ēē 1 w%‡GĔ÷úûOL*9*Ž@’D+" 9*ŽĒēĊ¬e%+ùY
E1F#$" +8”%‡h
E2F①#$" +8”%‡h
②#$" +8”%‡Y
é%%
E1F
3 0.000001 = 0.01h
3 0.001 = 0.1h
3 1 = 1h
3 1000 = 10h
3 1000000 = 100h
ËÞ߀
0.01G0.1G1 G10 G100h
E2F①
3 3000 =
3 1000 ×
3 3 = 10 × 1.442 = 14.42h
3 0.003 =
3 0.001 ×
3 3 = 0.1 × 1.442 = 0.1442h
ËÞ߀
14.42G0.1442h
②
3 0.000343 =
3 0.000001 ×
3 343 = 0.01 × 7 = 0.07
ËÞ߀
0.07Y
ð] 1-3— a!b ˜€d*Gc!d ˜€7š*G™ 𝑎2−𝑏2 +
3 𝑐𝑑=       
.
Þß―1
é% ĀTĕ
a+b=0Gcd=-1G
™ 𝑎2−𝑏2 +
3 𝑐𝑑= (𝑎+ 𝑏)(𝑎−𝑏) +
3 −1 =
0−1=-1.
ËÞße:-1.
【题型2  利用求平方根、立方根解方程】
– 2E25-26 ĖÔ¸·āĂ·ė·ĄąF’Â)*𝑎,𝑏G…†Hb,-“ ⊕”
𝑎⊕𝑏= 𝑎2−2𝑏YÂ𝑥+
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&(𝑥+ 1) ⊕ 8 = 0+%€            
Y
Þß𝑥1 = 3,𝑥2 = −5
>à ĀT…†+Ę,-GÚpÂ𝑥+HęĚu&‡G%r‡ĕ»Þß.
é%'T‡ĕ(𝑥+ 1)2−16 = 0G
∴(𝑥+ 1)2 = 16,
∴𝑥+ 1 = 4Š𝑥+ 1 = −4,
∴𝑥1 = 3,𝑥2 = −5Y
ËÞ߀
𝑥1 = 3,𝑥2 = −5.
Žěáâã+eHęĚu&%+ĜG)*+,-?/%…†+Ę,-e%+ù.
ð] 2-1E24-25 ÓÔ¸`·ĝôĞğ·õö÷øF`Úæ]N+ xY
(1)𝑥2 = 81h
(2)(𝑥+ 1)3 = −27Y
Þß(1)𝑥=± 9h
(2)𝑥= −4h
>àE1F Ā +…†%𝑥+(GHIˆ*^qI GĠġ˜€d*Y
E2F Ā" +…†%𝑥+(G´p−27+" Gµĕp𝑥+ 1+(G£ªp𝑥Y
áĎďâãĐ ?" +…†lD$YĔ÷úû ?" EfüHI*+"gÂ𝑎Gýþ
ÿI*JK𝑎+" F+|}e%+ùY
é%E1F%
𝑥2 = 81G 
𝑥=±
81
𝑥=± 9h
E2F%
(𝑥+ 1)3 = −27G 
𝑥+ 1 =
3 −27G
𝑥+ 1 = −3G
∴x = −4Y
ð] 2-2E24-25 ÓÔ¸`·ĢôģĤĥ·ØÙFêë&(2𝑥−1)2−16 = 9+ x +(€      
Y
Þß3 Š−2
>àáâãĐ#$ %&Y
´</ĕ»(2𝑥−1)2 = 25G™2𝑥−1 = 5Š2𝑥−1 = −5GĦ-‡Y
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é%%
∵(2𝑥−1)2−16 = 9
∴(2𝑥−1)2 = 25
∴2𝑥−1 = 5Š2𝑥−1 = −5
∴𝑥= 3Š𝑥= −2Y
ËÞ߀
−2Š 3Y
ð] 2-3E24-25 ÓÔ¸`·ħĨ·ØNF`Ú&N x +(
(1)2𝑥2 = 50h
(2)
1
3(𝑥+ 1)2 = 3h
(3)(2−𝑥)3 = 64h
(4)64(𝑥+ 1)3−125 = 0
Þß(1)𝑥= 5Š𝑥= −5
(2)𝑥= 2Š𝑥= −4
(3)𝑥= −2
(4)𝑥=
1
4
>àáĎďâãĐ#$ ?" ĩč*+(GĔ÷úû%HI*+ l" e%
+ùY
E1F Ā +…†%‡h
E2F Ā +…†%‡h
E3F#$" +…†%‡h
E4F#$" +…†%‡Y
é%E1F%
2𝑥2 = 50
𝑥2 = 25G
%ĕ
𝑥= 5Š𝑥= −5h
E2F%
1
3(𝑥+ 1)2 = 3
(𝑥+ 1)2 = 9G
𝑥+ 1 = 3Š𝑥+ 1 = −3
%ĕ
𝑥= 2Š𝑥= −4h
E3F%
(2−𝑥)3 = 64
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2−𝑥= 4G
%ĕ
𝑥= −2h
E4F%
64(𝑥+ 1)3−125 = 0
(𝑥+ 1)3 =
125
64 G
𝑥+ 1 =
5
4G
%ĕ
x = 1
4Y
【题型3  由平方根、立方根求值】
– 3E24-25 ÓÔ¸`·ÕĪī·õö÷øFČč|𝑎−6|@ 𝑎+ 2𝑏˜€d*G𝑐+ 5+" e 2G™
𝑎−2𝑏−𝑐+ €     Y
Þß ± 3
>àÆĎďâãĐd*+…†G" G GˆÊúûd…†e%ùY´#$" !
˜€d*+…†ĕp𝑎G𝑏G𝑐+(hĬĭ%ĕp𝑎−2𝑏−𝑐+(Gµ% ‡Y
é%%
∵|𝑎−6|@ 𝑎+ 2𝑏˜€d*G
∴|𝑎−6| +
𝑎+ 2𝑏= 0G
∴𝑎−6 = 0G𝑎+ 2𝑏= 0G
%ĕ
𝑎= 6G𝑏= −3G
∵𝑐+ 5+" e 2G
∴𝑐+ 5 = 8G
∴𝑐= 3
∴𝑎−2𝑏−𝑐= 6−2 × (−3)−3 = 9
∴𝑎−2𝑏−𝑐+ e ± 3Y
ËÞ߀
 ± 3Y
ð] 3-1Čč x +qI e𝑎+ 3@2𝑎−15G«2𝑏−1+-3 e 3Y
(1)𝑎,𝑏,𝑥+(h
(2)𝑎+ 𝑏−1+" Y
Þß(1)4G5G49
(2)2
>àáâã G-3 l" +…†YĔ÷úûM…†l8”e%+ùY
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E1F Ā @-3 +…†‡ĕ𝑎G𝑏+(Gµ%𝑥+(‡h
E2FĮ𝑎G𝑏+(Ĭĭ𝑎+ 𝑏−1NĦ-·#$" +…†‡ĕÞßY
é%E1F%
 ∵𝑥+qI e𝑎+ 3@2𝑎−15G«2𝑏−1+-3 e 3G
∴𝑎+ 3 + 2𝑎−15 = 0G2𝑏−1 = 9G
%ĕ
𝑎= 4G𝑏= 5h
∴𝑥= (𝑎+ 3)2 = 72 = 49h
E2F%
 ∵𝑎= 4G𝑏= 5G
∴𝑎+ 𝑏−1 = 4 + 5−1 = 8
ð] 3-2E24-25 ÓÔ¸`·įİıº·õö÷øFČč3𝑎+ 1+qI >Qe7−𝑚?2−2𝑚G9 + 𝑏+
" e 2Y
(1)𝑚G𝑎G𝑏+(h
(2) 7𝑎−𝑏+ Y
Þß(1)𝑚= 3G𝑎= 5G𝑏= −1
(2) ±
6
>àÆâã G" GĔ÷úûdčIJe%+ùh
E1F Ā …†G" …†Úĕ7−𝑚+ 2−2𝑚= 0G9 + 𝑏= 23G‡p𝑚G𝑎G𝑏+(h
E2F´p7𝑎−𝑏+(Gµ#$ …†pÞߝ‡Y
é%E1F%
tTĕG7−𝑚+ 2−2𝑚= 0 
%ĕ𝑚= 3        
Ë7−𝑚= 4,2−2𝑚= −4
∴3𝑎+ 1 = 42G
%ĕ𝑎= 5 
'T9 + 𝑏= 23G
%ĕ𝑏= −1h
E2F∵7𝑎−𝑏= 7 × 5−(−1) = 36G
36 = 6 G6 + € ±
6 G
ij_ 7𝑎−𝑏+ € ±
6Y
ð] 3-3E24-25 ÓÔ¸`·ĴİĘĵ·ØÙFČč2𝑎+ 1+ e ± 3.3𝑎+ 2𝑏−4+" e−2G
(1)𝑎?𝑏+(h
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(2)3𝑎+ 𝑏+ 
Þß(1)𝑎= 4G𝑏= −8
(2) ± 2
>àáĎďâãĐ" G +…†G Ā" G +…†%‡Y
E1F'" G +…†‡č2𝑎+ 1 = 9G3𝑎+ 2𝑏−4 = −8G®·‡p aGb +(Y
E2FĶ aGb +(Ĭĭ3𝑎+ 𝑏Gµ ‡Y
é%E1F%
 ∵2𝑎+ 1+ e ± 3G3𝑎+ 2𝑏−4+" e−2G
∴2𝑎+ 1 = 9G3𝑎+ 2𝑏−4 = −8G
∴𝑎= 4G
∴𝑏= −8Y
E2F%
∵3𝑎+ 𝑏= 3 × 4−8 = 4G
∴3𝑎+ 2𝑏+ € ± 2Y
【题型4  实数的运算】
– 4E24-25 ÓÔ¸`·ķĸĹĺ·ØNFĮ 1G 2G 3G 6„fĻ]ļÚG—Ċ…(𝑚,𝑛)yzĽ𝑚ļ¹
ºđ¼Ľ𝑛I*G™(5,4)@(12,4)yz+q*rľe         
.
Þß0
>àÆĎďâãĐ*Ŀ+ðŀĊ¬G)*+,-GŁƒ*ĿðŀĊ¬eùY
 Ā*+ļÚÀ‡čGĽHļ
1 I*GĽĚļ 2 I*YĽłļ 3 I*GĽķļ 4 I*G…Ľ𝑚ļ^𝑚I
*G¹ĽHļ»𝑚ļŃ^
1 + 2 + 3 + 4 + … + 𝑚=
𝑚(𝑚+1)
2
I*G Ā*+ļÚÀG‹ķI*HIńŅG
 ĀņTŇŁpĽ𝑚ļĽ𝑛I*»ňeʼnI*·µĦ-Y
é%%
 Ā*+ļÚÀ‡čGĽHļ
1 I*GĽĚļ 2 I*YĽłļ 3 I*GĽķļ 4 I*G…
Ľ𝑚ļ^𝑚I*G¹ĽHļ»𝑚ļŃ^
1 + 2 + 3 + 4 + … + 𝑚=
𝑚(𝑚+1)
2
I*G Ā*+ļÚÀG‹ķI
*HIńŅG
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(5,4)yzĽ 5 ļ¹ºđ¼Ľ 4 I*e 2G
∵Ŋ 11 ļŃ^1 + 2 + 3 + 4 + … + 11 =
11×(11+1)
2
= 66 (I)*G
∴ (12,4)yzĽ 12 ļĽ 4 I*Ľ 70 I*G
∵70 ÷ 4 = 17⋯2G
∴ (12,4)yz+*e 2G
∴ (5,4)@(12,4)yz+q*rľe 2−2 = 0G
ËÞ߀
0Y
ð] 4-1E2025 ŋÔ¸·āĂ·Ō÷øFĦ-
 16−(−1)2025−
3 27−|1−2|Y
Þß3−2
>àáĎďâãĐ)*+,-G´Ħ--3 ?" GōĦ-œ’(?§G¶·Ħ-¥¦
À‡ĕ»ÞßY
é%%
Ŏ] = 4−(−1)−3−( 2−1)
= 4 + 1−3−2 + 1
= 3−2Y
ð] 4-2E24-25 ŋÔ¸·ĝôŏĺ·ØÙFĊ…
[𝑚]yzVՑ𝑚+¶<*G{𝑚}yz𝑚+9*=>G
[𝑚] + {𝑚} = 𝑚GMN𝑚€)*Y–f[3.14] = 3G{3.14} = 0.14G[3.14] + {3.14} = 3.14YĦ-
[−3]−{ 5}
=       
Y
Þß−5
>àáâãĐʅ†`+)*,-G./*+0-G/%ʅ†e%+ùY>Q0Ħ−3? 5
žʼnqI<*rsGµ Āʅ†ĕp[−3] = −2G{ 5} =
5−2Gqőd¦‡ĕpÞßY
é%%
∵−2 < −3 < −1G2 <
5 < 3G
∴[−3] = −2G{ 5} =
5−2G
∴[−3]−{ 5} = −2−( 5−2) = −5Y
ËÞ߀
−5Y
ð] 4-3E24-25 ÓÔ¸·ŒĴœò·ØNF‹I&²ö'—ŔŕŖŗŘ{GřŚĦĐHö,-&²f
Ļ
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–f
Îśĭ x +(€−1ÐGĦ-Ñü 3 < 4hĮśĭ(ð€(−1) +1Ŝ0GĦ-Ñü€ 6 < 4hµĮśĭ
(ð€Đ0 + 1 = 1GŝŞ,-G»Ħ-ÑüV9 4GşśpŠÑüY
šŇâ`ÚĜY
(1)Îśĭ x +(€ 5G™śp y +(evŢţšÚ]Ħ-Y
(2)ÎŤťśĭ x +(€ 1GšÉ‘Ħ-ÛŦŧ‘au&²,¤·şxśp yY
Þß(1) 21
(2)4 u
>àáâãĐ)*+,-G/%TGúûŨĻN+,-À™e%+ùY
E1F ĀŨĻN+,-&²Ħ-‡h
E2F ĀŨĻN+,-&²Ħ-G»ÑüŠg 4 śpÑü€ũY
é%E1FÎśĭ x +(€ 5 ÐG
™^5−(−2) = 7G7 × 3 = 21G
« 21 > 4G
∴śp y +(e 21Y
E2FÎśĭ x +(€ 1 ÐG
™^1−(−2) = 3G3 × 3 = 9G 9 < 4GŝŞĦ-h
ĽĚuśĭ x +(€1 + 1 = 2ÐG
™^2−(−2) = 4G4 × 3 = 12G 12 < 4GŝŞĦ-h
Ľłuśĭ x +(€2 + 1 = 3ÐG
™^3−(−2) = 5G5 × 3 = 15G 15 < 4GŝŞĦ-h
Ľķuśĭ x +(€3 + 1 = 4ÐG
™^4−(−2) = 6G6 × 3 = 18G 18 > 4Gśp𝑦h
ij_ŧ‘ 4 u&²,¤·şxśp yY
【题型5  无理数的估算】
– 5E24-25 ÓÔ¸`·ŪūŬŭ·ØÙFŮ®eů+ŚĦGŰeHű99+ŲųGŴŵnŶ“ů
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”YfĻG
𝐵𝐶
𝐴𝐵+(ÌŷŸ:
5−1
2 G™`Ú0-ˆÊ+eE    F
AY0 <
5−1
2
<
2
5
BY
2
5 <
5−1
2
<
1
2
CY
1
2 <
5−1
2
< 1
DY1 <
5−1
2
< 2
ÞßC
>àáâãĐ./*+0-Gáâã./*+0ĦGVg]+8”GˆÊŹź+Ÿ e%á
+ùY'Tč4 < 5 < 9G™2 <
5 < 3G ĀVg]+8”Ħ-%‡Y
é%%
'TčG 4 < 5 < 9G
∴2 <
5 < 3G
∴1 <
5−1 < 2G
∴
1
2 <
5−1
2
< 1Y
Ëç
CY
ð] 5-1E24-25 ŋÔ¸`·ĴŻżŽ·õö÷øFČč432 = 1849G442 = 1936G452 = 2025G462
= 2116Y—𝑛€<*«𝑛<
2022 < 𝑛+ 1G™𝑛+(€E   F
AY43
BY44
CY45
DY46
ÞßB
>àáĎďâã./*+0-GĔ÷úû./*+0-e%+ùh´ ĀŔN+*Ā0- 2022
+9G£ªĜ‡%Y
é%%
'T‡č
1936 < 2022 < 2025
∴44 <
2022 < 45G
∴𝑛= 44h
Ëç BY
ð] 5-2E24-25 ĖÔ¸·ĴŻżŽ·ØNF—𝑎<
23 < 𝑏G«𝑎!𝑏€žŞˆ<*G™𝑎+ 𝑏=        
Þß9
>àáâã)*+0-@9:;+xſG´0-p 23+ƀ(Ÿ Gĕp𝑎G𝑏+(G£ª‡ĕpÑ
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ÒY ĀTp𝑎G𝑏+(e%+ùY
é%%
∵16 < 23 < 25G
∴4 <
23 < 5G
∵𝑎G𝑏€qIžŞ<*G
∴𝑎= 4G𝑏= 5G
∴𝑎+ 𝑏= 4 + 5 = 9Y
ËÞ߀
9Y
ð] 5-3E24-25 ÓÔ¸`·ĢƁƂÕ·ØÙFČč𝑘<
17−1G«𝑘€ˆ<*G™𝑘+(‡_e           
EopHI‡FY
Þß1EÞßVƃHF
>àáĎďâãĐ./*+0-!Vg]+%gčIJŽGúû./*+0-Àe%+ùY
´0- 17ĕƀ(Ÿ GµÊ… k +ƀ(Ÿ G®· ĀVg]%+…†‡%ÞY
é%%
∵16 <
17 <
25e<*h
∴3 <
17−1 < 4eƄā*h
∴𝑘≤3G
∵𝑘€ˆ<*G
∴𝑘= 1Š 2 Š 3Y
ËÞ߀
1EÞßVƃHFY
【拔尖篇】
【题型6  算术平方根的双重非负性】
– 6E24-25 ÓÔ¸`·ĝƁƅƆ·õö÷øFČč aGb €)*G«𝑏=
𝑎−8−8−𝑎+25G™
3 𝑎+
𝑏+
(€E    FY
AY10
BY9
CY8
DY7
ÞßD
>àáâã-3 {"+ŕƇG" G Ā-3 +L*e67*ĕ»𝑎= 8G£ª
ĕ𝑏= 25G¶·Ĭĭ
3 𝑎+
𝑏Ħ-‡Y
é%%
∵𝑏=
𝑎−8−8−𝑎+25G𝑎−8 ≥0G8−𝑎≥0G
∴𝑎= 8G𝑏= 25G
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∴
3 𝑎+
𝑏=
3 8 +
25 = 2 + 5 = 7G
Ëç
DY
ð] 6-1E25-26 ŋÔ¸·āĂ·ƈƉ÷øF—𝑎+
𝑎−3 = 3G 𝑎+ 3+(Y
Þß 6
>àáĎďâã+eĚu ]GĔ÷úûĚu ]+8”e%+ùY
´ĮŎ]ŀ€ 𝑎−3 = 3−𝑎Gµ$Vg]Ř%‡Y
é%%
ï€𝑎+
𝑎−3 = 3G
ij_ 𝑎−3 = 3−𝑎G
ij_𝑎−3 ≥0,3−𝑎≥0G
ij_𝑎= 3G
ij_ 𝑎+ 3 =
3 + 3 =
6Y
ð] 6-2E24-25 ÓÔ¸`·ñòñĺ·ØNFČč|𝑎−3| +
3 + 𝑏= 0G™
𝑎
𝑏
2025
= E    F
AY1
BY−1
CY2
DY−2
ÞßB
>àáâ㜒(G-3 G^/*+§G%+ùep𝑎?𝑏+(Y
 Āœ’(?-3 +678G%ĕ𝑎?𝑏+(GĬĭĦ-‡Y
é%%
|𝑎−3| +
3 + 𝑏= 0G|𝑎−3| ≥0G 3 + 𝑏≥0G
∴|𝑎−3| = 0G 3 + 𝑏= 0G
∴𝑎−3 = 0G3 + 𝑏= 0G
∴𝑎= 3G𝑏= −3G
∴
𝑎
𝑏= −1
∴
𝑎
𝑏
2025
= (−1)2025 = −1
Ëç
BY
ð] 6-3E2025 ŋÔ¸`·ĝƁ·Ō÷øFČč6©)* aGb êë|2𝑎−4| + |𝑏+ 2| +
(𝑎−4)𝑏2
+4 = 2𝑎G™𝑎+ 𝑏=      
Y
Þß2
>àáĎďâãĐ-3 +678Gœ’(+678GĮ]Ɗð\€|2𝑎−4| + |𝑏+ 2| +
(𝑎−4)𝑏2
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= 2𝑎−4G'-3 +678Gœ’(+678ĕp2𝑎−4 ≥0GµŀƋ]Ɗ‡ĕp|𝑏+ 2| +
(𝑎−4)𝑏2
= 0Gµ Ā-3 +678Gœ’(+678‡ĕp𝑏= −2G𝑎= 4G£ªĬĭĬ*]‡ĕpÞ
ßY
é%%
∵|2𝑎−4| + |𝑏+ 2| +
(𝑎−4)𝑏2 +4 = 2𝑎G
∴|2𝑎−4| + |𝑏+ 2| +
(𝑎−4)𝑏2 = 2𝑎−4G
∴2𝑎−4 ≥0G
∴2𝑎−4 + |𝑏+ 2| +
(𝑎−4)𝑏2 = 2𝑎−4G
∴|𝑏+ 2| +
(𝑎−4)𝑏2 = 0
∴𝑏+ 2 = 0G𝑎−4 = 0G
%ĕG𝑏= −2G𝑎= 4G
™𝑎+ 𝑏= 4−2 = 2G
ËÞ߀
2Y
【题型7  实数的大小比较】
– 7E24-25 ÓÔ¸`·ħĨƌƍĘ·ØNF:;9
 11     3h−5     −2h
5+1
2     
3
2YEĆ“ > ”!
“ = ”Š“ < ”FY
Þß     
>     
<     
>
>àáâãĐ)*+9:;GĔ÷úû_čIJŽe%Þá+ùY
 Ā)*+9:;À%ޝ‡Y
é%%
 ∵ 
11 >
9G
∴
11 > 3h
∵5 >
4 = 2G
∴−5 < −4G
∴−5 < −2h
∵5 >
4 = 2G
∴
5 +1 > 3G
∴
5+1
2
>
3
2h
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ËÞ߀
 > G < G > Y
ð] 7-1E24-25 ÓÔ¸`·ÕƎò·ØNF—𝑎=
3 7G𝑏=
5G𝑐= 2G™ aGbGc +9Á€E    F
AY𝑏≤𝑐< 𝑎
BY𝑏< 𝑎< 𝑐
CY𝑎< 𝑐< 𝑏
DY𝑎< 𝑏< 𝑐
ÞßC
>àáâã)*+9:;G´#$ƏƐÀ0- aGb +(Gµ:;9‡Y
é%%
∵1 < 7 < 8G4 < 5 < 9G
∴
3 1 <
3 7 <
3 8G 4 <
5 <
9G
1 <
3 7 < 2G2 <
5 < 3G
∴1 < 𝑎< 2G2 < 𝑏< 3G
Ƒ𝑐= 2G
∴𝑎< 𝑐< 𝑏G
Ëç
CY
ð] 7-2E25-26 ŋÔ¸·āĂ·ƈƉ÷øF:; 65−2@ 10 +2+9Y
Þß 65−2 >
10 +2
>àÆâã./*+0-?)*+9:;G‡É‘0-qI./*+Ÿ G£ªÊ…qIĬ*]
+Ÿ Gµ£¤:;Y
é%%
ï€8 <
65 < 9G3 <
10 < 4G
ij_6 <
65−2 < 7G5 <
10 +2 < 6G
ij_ 65−2 >
10 +2Y
ð] 7-3E25-26 ŋÔ¸·āĂ·ƈƉ÷øF:;`ÚæŘ*+9
(1)
1
2@
10
6 h
(2)
6−1
2 @ 1Y
Þß(1)
10
6 >
1
2
(2)
6−1
2
< 1
>àáâã)*:;9GĔ÷úû)*:;9+ÀGe%+ù
E1F0-À:;9‡h
E2F0-À:;9‡Y
é%E1F%
∵10 >
9 = 3G
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∴
10
6 >
3
6G
10
6 >
1
2h
E2F∵4 <
6 <
9G
∴2 <
6 < 3G
∴1 <
6−1 < 2G
∴
6−1
2
<
2
2G
6−1
2
< 1Y
【题型8  求整数部分和小数部分】
– 8E24-25 ŋÔ¸·ĢôĘƒ·õö÷øF— 3+<*=>€ aG9*=>€ bG4−3+<*=>€
cG9*=>€ dG™
𝑏+𝑑
𝑎𝑐+(€E   F
AY
1
2
BY
1
4
CY
3−1
2
DY
3+1
2
ÞßA
>àáâãĐ0-./*+9Gɑ0- 3žʼnqI<*rsG¹ªÊ…𝑎= 1G𝑏=
3−1GÉ
‘0-4−3žʼnqI<*rsG¹ªÊ…𝑐= 2G𝑑= 2−3G®·Ķ a!b!c!d Ĭĭĕ*(Y
é%%
∵3+<*=>€ aG9*=>€ bG1 <
3 < 2G
∴𝑎= 1G𝑏=
3−1G
∵1 <
3 < 2G
∴−2 < −3 < −1G
∴2 < 4−3 < 3G
∵4−3+<*=>€ cG9*=>€ dG
∴𝑐= 2G𝑑= 4−3−2 = 2−3G
∴
𝑏+𝑑
𝑎𝑐=
3−1+2−3
1×2
=
1
2Y
Ëç
AY
ð] 8-1E24-25 ÓÔ¸`·įįò·ØNF—6 +
5+<*=>e𝑚G9*=>e𝑛G™|𝑛−𝑚|€E   F
AY 5−10
BY10−5
CY 5−6
DY6−5
ÞßB
>àáâã@./*+<*=>^+Ħ-G)*+,-GƏƐÀp𝑚,𝑛+(GµĬ(Ħ-‡Y
é%%
∵4 <
5 <
9G
∴2 <
5 < 3G
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∴8 < 6 +
5 < 9G
∴𝑚= 8G𝑛= 6 +
5−8 =
5−2G
∴|𝑛−𝑚| = | 5−2−8| = 10−5h
Ëç BY
ð] 8-2E24-25 ŋÔ¸·ĝôŏĺ·ØÙFĊ…
[𝑚]yzVՑ𝑚+¶<*G{𝑚}yz𝑚+9*=>G
[𝑚] + {𝑚} = 𝑚GMN𝑚€)*Y–f[3.14] = 3G{3.14} = 0.14G[3.14] + {3.14} = 3.14YĦ-
[−3]−{ 5}
=       
Y
Þß−5
>àáâãĐʅ†`+)*,-G./*+0-G/%ʅ†e%+ùY>Q0Ħ−3? 5
žʼnqI<*rsGµ Āʅ†ĕp[−3] = −2G{ 5} =
5−2Gqőd¦‡ĕpÞßY
é%%
∵−2 < −3 < −1G2 <
5 < 3G
∴[−3] = −2G{ 5} =
5−2G
∴[−3]−{ 5} = −2−( 5−2) = −5Y
ËÞ߀
−5Y
ð] 8-3E24-25 ÓÔ¸`·įĴƓ·õö÷øFČč)*8 +
13+<*=>€𝑎G9*=>e𝑚h)*
9−13+<*=>€𝑏G9*=>e𝑛Y
(1)op𝑎G𝑚G𝑏G𝑛+(h
(2)12𝑚+ 12𝑛+
𝑎+ 𝑏−7+(+ h
(3)
3 (𝑚+ 𝑛)2025 +(𝑎−2𝑏)+ 6+(Y
Þß(1)𝑎= 11G𝑚=
13−3G𝑏= 5G𝑛= 4−13
(2) ± 3
(3)2
>àáâãĐ./*+0-! !" GĔ÷úû_čIJŽōƔƕ,$e%Æ+ùY
E1F0-ĕp3 <
13 < 4G¹ª‡ĕ11 < 8 +
13 < 12G5 < 9−13 < 6GÑCT‡ĕ%h
E2FĮE1FN𝑎= 11G𝑚=
13−3G𝑏= 5G𝑛= 4−13Ĭĭij]Ɗ£¤Ħ-GµÑC +…†Ħ
-‡ĕ%h
E3FĮE1FN𝑎= 11G𝑚=
13−3G𝑏= 5G𝑛= 4−13Ĭĭij]Ɗ£¤Ħ-GµÑC" +…†Ħ
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-‡ĕ%Y
é%E1F%
∵9 < 13 < 16G
∴9 <
13 <
16G3 <
13 < 4G
∴11 < 8 +
13 < 12G5 < 9−13 < 6G
∵)*8 +
13+<*=>€𝑎G9*=>e𝑚G)*9−13+<*=>€𝑏G9*=>e𝑛G
∴𝑎= 11G𝑚= 8 +
13−11 =
13−3G𝑏= 5G𝑛= 9−13−5 = 4−13h
E2F%
'E1F‡ĕ
𝑎= 11G𝑚=
13−3G𝑏= 5G𝑛= 4−13G
12𝑚+ 12𝑛+
𝑎+ 𝑏−7
= 12 × ( 13−3) + 12 × (4−13) +
11 + 5−7
= 12 13−36 + 48−12 13 + 4−7
= 9G
∴12𝑚+ 12𝑛+
𝑎+ 𝑏−7+(+ € ±
9 =± 3h
E3F%
3 (𝑚+ 𝑛)2025 +(𝑎−2𝑏)+ 6
=
3 ( 13−3 + 4−13)2025 + (11−2 × 5) + 6
=
3 1 + 1 + 6
=
3 8
= 2Y
【题型9  实数与数轴综合运用】
– 9E24-25 ÓÔ¸`·ĢƁƖƗƘ·ØÙFƙĖƚ-3ƛNƜƝ3ƞ
“ƜƝæ٧GōªL¨rG
Ɵ”G𝑐=
𝑎2 + 𝑏2E𝑎€“Ɯ”G𝑏€“Ɲ”G𝑐€“Ɵ”F—“Ɯ”€2G“Ɲ”€3G™“Ɵ”žfĻijz*A‡yz
žE    
F
AY𝐴Ž
BY𝐵Ž
CY𝐶Ž
DY𝐷Ž
ÞßC
>àáâã./*+0-G)*@*AG/%TōÚĕˆÊ+-]e%+ùY ĀTÚ]
Ħ-·0-M9G®·Ê…Mž*A+wƠ‡Y
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é%%
—“Ɯ”€2G“Ɲ”€3G™ 22 + 32 =
13G
∵9 < 13 < 16G
∴3 <
13 < 4G
™“Ɵ”žfĻijz*A‡yzž𝐶ŽG
Ëç
CY
ð] 9-1E24-25 ÓÔ¸`·ơ×ƢÕ·ØNFfĻGˆ\𝐴𝐵𝐶𝐷+¿ƣ€ 7GƤŽ A ž*Ayz+*
€−1G—Ž E ž*AEŽ E žŽ A +ºƥFG«𝐴𝐷= 𝐴𝐸G™Ž E ijyz+*€          
Y
Þß−7−1
>àáâã-3 G)*@*AG´pˆ\+ÄĞG£ª ĀqŽs+ƦƧpŽ E ijy
z+*‡Y
é%%
∵ˆ\𝐴𝐵𝐶𝐷+¿ƣ€ 7G
∴𝐴𝐸= 𝐴𝐷=
7G
∵ƤŽ A ž*Ayz+*€−1G
∴Ž E ijyz+*€−7−1h
ËÞ߀
−7−1Y
ð] 9-2E24-25 ŋÔ¸`·ƨŬĴƩƪĴ·ØÙF)*𝑎G𝑏ž*A+wƠfĻijzGŀƋ
|𝑎+ 1|−(𝑏−1)2
+
(𝑎−𝑏)2 =           
Y
  
Þß2
>àáâãĐ)*@*AG#$*Aĕp−1 < 𝑎< 0G1 < 𝑏< 2G£ªŀƋ‡Y
é%%
'*AGĕ−1 < 𝑎< 0G1 < 𝑏< 2G
∴𝑎+ 1 > 0G𝑏−1 > 0G𝑎−𝑏< 0G
∴Ŏ] = 𝑎+ 1−|𝑏−1| + |𝑎−𝑏|
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= 𝑎+ 1−(𝑏−1)−(𝑎−𝑏)
= 𝑎+ 1−𝑏+ 1−𝑎+ 𝑏
= 2G
ËÞ߀
2Y
ð] 9-3E24-25 ÓÔ¸`·ĝôZƎ·ØNFfĻGɑƫÄĞ€ 1 +ˆ\GƬxƒÊ+Ķ 2yzž*
AŽ𝐴1ƭGƮ𝐴1¼ƥ¶Ì+<*Ž€𝐵1G_Ž𝐵1€iƯG𝐴1𝐵1€ưƱƫưiGƲ*AŽ𝐴2GƮ𝐴2¼ƥ
¶Ì+<*Ž€𝐵2G_Ž𝐵2€iƯG𝐴2𝐵2€ưƱƫưiGƲ*AŽ𝐴3GfÆŝŞG™𝐴2026𝐵2026+Ğ
€          
Y
Þß 2−1/−1 +
2
>àáâãĐ)*+,-+Ċ¬G)*@*AG´pŽ B yz+*ĕ»𝐴1𝐵1 = 2−2G™𝐴2yz
+*€4−2Gµp𝐵2yz+*€3G™𝐴2𝐵2 =
2−1G®·tuyz𝐴3𝐵3 = 2−2G𝐴4𝐵4 =
2−1G𝐴5𝐵5
= 2−2h𝐴6𝐵6 =
2−1h‡Ł»Ċ¬%Y
é%%
'TĕGŽ𝐴1yz+*€ 2G
∵1 <
2 <
4G
∴1 <
2 < 2G
∴𝐵1yz+*€ 2G
G∴𝐴1𝐵1 = 2−2G™𝐴2yz+*€2 + 2−2 = 4−2G
∵1 <
2 < 2G
∴−2 < −2 < −1
∴2 < 4−2 < 3G
∴𝐵2yz+*€3G
∴𝐴2𝐵2 = 3−(4−2) =
2−1G
³/‡ĕ𝐴3𝐵3 = 2−2h
𝐴4𝐵4 =
2−1h
𝐴5𝐵5 = 2−2h
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𝐴6𝐵6 =
2−1h
……G
_ƁƳ‡ĕGÎ n €ƴ*Ð𝐴𝑛𝐵𝑛= 2−2GÎ n €Ƶ*Ð𝐴𝑛𝐵𝑛=
2−1G
∴𝐴2026𝐵2026 =
2−1
ËÞ߀
 2−1Y
【题型10  实数运算的应用】
– 10E24-25 ÓÔ¸`·ĝôĞğ·õö÷øFƶƷ^HIƸƣ€200cm3+ƹƺŎƻGƶƷƼďĮÿ
Iƹƺƽŀō5Ęƾƿ{Ę+\ǀY
(1)—ĮŎƻ5Ęƾƿ{HIň¿€ˆ\!ǁ€8cm+ĞƸGĞƸň¿ˆ\+ÄĞY
(2)ƶƷĉĮŎƻƾƿ{łI9d³+ˆƸƹƺGǂĄƄ{·ǃ`+ĵƻƸƣ€8cm3GǂĄ{+‹
HI9ˆƸƹƺ+DŽĞY
Þß(1)5cm
(2)4cm
>àáâã-3 G" +D$G
E1F ĀĞƸƸƣ+Ħ-]“ĞƸ+Ƹƣ = ň¿ƣ × ǁ”Ú&%‡h
E2F Ā“ˆƸƸƣ+Ħ-À_l3I9ˆƸƸƣ@ÅƸƣrs+Á”Ú&%‡h
/%-3 !" +…†eˆÊ%Þ+ùY
é%E1F%
ŚĞƸň¿ˆ\+ÄĞ€𝑥cmG
tTGĕ
8𝑥2 = 200G
%ĕ
𝑥= 5Š𝑥= −5E7(VRCTGDždžFG
Þ
ĞƸň¿ˆ\+ÄĞ€5cmh
E2F%
Ś‹HI9ˆƸƹƺ+DŽĞ€𝑦cmG
tTGĕ
3𝑦3 +8 = 200G
%ĕ
𝑦= 4G
Þ
‹HI9ˆƸƹƺ+DŽĞ€4cmY
ð] 10-1E24-25 ÓÔ¸·ŒĴ·ØNFfĻGHIň¿ưƱ€3cm+LJƊ¡LjŶHmljNJYÎLJƊˆ
NjÐGLJ¡ljNJ+ǁnj€16cmhšNjÐGċĵ=>+ǁnj€4cmYLJ¡+ljNJˆǍšê 2 IHǎ+ˆ
ƸǏǐEπƀ 3GǏǐ+ǑnjVĦFY
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(1)ŠLJƊ+ǏƣELjêÐljNJ+ƸƣFevŢ"ǒǓţ
(2)ˆƸǏǐ+DŽĞevŢǒǓţ
Þß(1)540cm3
(2)6cm
>àáâã^/*+°C,-!HI*+" GÈǔlZ[a›Ƹ+ƸƣGǕǖTGĕp“LJ
Ɗ+Ǐƣ@³ň!ǁ€(16 + 4)cm+iǗƸƣdg”e%+ùY
E1FLJƊ+Ǐƣ@³ň!ǁ€(16 + 4)cm+iǗƸƣdgG'Ƈ%h
E2Fǘ´pLJ¡+ljNJ+ƸƣG®· ĀLJ¡+ljNJˆǍšê 2 IHǎ+ˆƸǏǐ%‡Y
é%E1F%
𝑉= π𝑟2ℎ≈3 × 9 × (16 + 4) = 27 × 20 = 540(cm3)h
E2F%
ï€𝑉溶液= π𝑟2ℎ≈3 × 32 × 16 = 432(cm3)Y
ij_DŽĞ =
3 432 ÷ 2 = 6cmY
ð] 10-2ǙHǚǛė·G€ĐyǜžăØtǝǞǟƄ{*Ąą+9Ǡ?9ǡ³G*řǢ…
žǣÇǤǥŽ 2 ǦǧǨŒe 16 ę+ǩǪǫǬǭġY¹ǩǪĥ»Ǜ+‹ǮǯǧǰDZDzŒ€ 1.6 ęG'
Â*řeŠǥ+dzǴGïƋǧŒ‡_Ű$HIǥǵDZ+ 5 ęǥǶǷ’‹Ǯǯǧ+ÅǨ¦Ǹ 5
ęEǯǧÅǨVnǰDZDzG³HǯǧǹǺǻŰ$HIǶǷFYǼ ĀŠǩǪĥ+ǽƕēGÎǯǧÅǨEV
nǰDZDzFê 30 ęÐG5 ę+ǥǶǷ‡Ǿǿ€HI 7 ę+ĥȀǶǷG‡_ȁǯǧÅǨEVnǰDZDzF
¦Ǹ 7 ęÎ*ř³ÐŽĐ 2 ǦǩǪƒȂ`ǧȃȄÐG9Ǡ³Û
“řGȅġ‡_ǿHb`ǧ]G
ǽȆvȇ”š³ġ>à9Ǡ³+`ǧ]GōĦ-páuÇǤÅDz$EǷnǰDZDzF¶Ȉ‡€        
ęY
Þß25.2
>à>QĦ-qb`ǧ+]G:;ʼnHbÅDz$ȆȈ‡Y
é%ĽHb`ǧ]€ȉȊqǦǩǪ
CĦDz$€
16 + 16 + 1.6−7 = 26.6ę
ĽĚb`ǧ]€`qIǯǧG‹IǯǧȊHǦǩǪ
CĦDz$€
(16 + 1.6−5) × 2 = 25.2ę
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ËçȋĽĚbȆȌ-G¶ȈDz$€ 25.2 ę
ËÞ߀
25.2Y
ŽěáâãĐ)*,-+)ȍD$G>ȎÒe%+ùY
ð] 10-3ǣ9PƒȂLĈHƺĞ€32mGȏ€21m+Ğ\ċȐG
(1)ßH
fĻGĮÿƺċȐbȑȒGNsȓHŕȔȕ+9ȖG9Ȗ+ºÄȗđ¼Ē1mƬeĠ+¼Ä
ȗY™ÿƺȑȐ+¿ƣ€ _____ m2h
(2)ßĚ
ȓūHIĞeȏ+ 1.5 ȘG¿ƣ€384m2+șȚțG—:Ȝ$+șȚțďО23m»30mr
sGȏž13m»20mrsY ÿIșȚțx$K:Ȝȝţ ōÛŦ/'Y
Þß(1)651
(2)xG/'[%à
>àáâãĐĻ\+ĒG +…†gčIJY
E1F'TGȑȐ+Ц91mGȏVðG慎ĕȑȐ+¿ƣh
E2FŚȏ𝑥mG™Ğ€1.5𝑥mG Ā¿ƣ]‡ĕ x +&G' +…†‡ĕ xGµ’ x +
(£¤0-G—êëT‡GȞ™V¤Y
é%E1F%
'TG9Ȗ+ºÄȗđ¼Ē1mƬeĠ+¼Äȗ9Ȗ+ȏ€1mG
™ȑȐ+Ц91mGȏVðG
¿ƣ€21 × (32−1) = 651m2h
ËÞ߀
651Y
E2FxG/'f`
Śȏ𝑥mG™Ğ€1.5𝑥mG
tT^
1.5𝑥2 = 384G 𝑥2 = 256
∵𝑥> 0G
∴𝑥= 16G1.5𝑥= 24
RCО23m»30mrsGȏž13m»20mrsG
∴ÿIșȚțx$K:ȜY
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