第十九章实数(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(沪教)

第十九章  实数（举一反三讲义）全章题型归纳
 2024
 ...................................................................................................................................................................5
 1   !"# ! ........................................................................................................................5
 2  $% !"# !& ' ............................................................................................................6
 3  ( !"# !) ........................................................................................................................9
 4  *+,-. ..........................................................................................................................................11
 5  /0+,1. ......................................................................................................................................13
23 .................................................................................................................................................................15
 6  .4 !,56789 ..................................................................................................................15
 7  *+,:;<= ..................................................................................................................................17
 8  >+?@A;+?@ ......................................................................................................................19
 9  *+B+CDE-% ..........................................................................................................................21
 10 *+-.,F% ..................................................................................................................................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|= 𝒂G𝒂≥𝟎H
−𝒂G𝒂< 𝟎H
62=|6|=6
(−6)2=|−6|=6
𝒂𝟐
( 𝐚)𝟐 = 𝐚G𝐚≥𝟎H
( 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=aI 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.JK+ a ,# !,-.LMN# IOP a LMQN +.
2.N AN# ,RS
N 
N# 
-.T	
±
3
QN +
:,
UV+
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K+
0 "#
+

#$+,-#
efaG"#


知识点6  无理数
/WXYZ;+LM/0+[
/0+,\]^_`abcd
G1HN NXe,+,fF !g/0+Ih 2I
3 5ij
G2HklmπnJop`π,+,h 2πI
π
2ij
G3Ha/WXYZ;+^_qr,+,h 0.101 001 000 1…GsK 1 tuvwxJK 0)i[V/0+,;
+?@y+/Wj/0+,;+?@XYZj/0+Xz{|}@+,^_[
知识点7  实数的概念及分类
1. ~
`0+A/0+€‚*+[
2. @ƒ
*+`sd@ƒ„…
G1H†‡ˆ@ƒ
*+‰@‚`0+A/0+[
G2H†Š89@ƒ
*+‰@‚Š*+"0"8*+[
知识点8  实数与数轴的关系
Š>+
Š@+
8>+
8@+
*+
Š*+
8*+
Š`0+
Š/0+
8`0+
8/0+
0
*+
`0+
0
/0+
Š`0+
8`0+
Š/0+
8/0+
`W;+ŒYZ;+
/WXYZ;+
Š>+
Š@+
8>+
8@+
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JK*+Ž‰a%+C,JK‘{|j’“‘I+C,JKŽ{|JK*+[*+B+C
,JJ”F.
知识点9  实数范围内的有关概念
•
9–
—˜
f’+
™ a B b š‚f’+I›𝑎+ 𝑏= 0
3,f’+g−3
œ+
™ a B b š‚œ+I›𝑎𝑏= 1
2 ,œ+g
1
2
U*+,ž”)Žg78+IŸ|𝑎| ≥0
|𝑎|
𝑎(𝑎> 0)
0(𝑎= 0)
−𝑎(𝑎< 0)
ž”)
š‚f’+,sK+,ž”)fiIŸ|𝑎| = |−𝑎|
| 2| = |−2| =
2
知识点10  实数的运算
 *+¡¢£IX¤‰a¥¦§"¨"©"ªGª+X‚«H"© -.I¬­‰a¥¦N# -.an
78*+,N -.[
`0+,-.9–A-.® *+¡¢£¯°±%I*+²E-.,³´B`0+²E-.,³´fµI
¶© "N I·©ªI¸¹§¨[µº-.†»¼½¾,³´¥¦I`¿	,¶.¿	ÀÁ,[
知识点11  实数的大小比较
`0+:;<=,  *+¡¢£¯°±%[
s*+,:;Ãhb
Š*+Ž:Ä 0I8*+Ž;Ä 0IŠ*+:ÄJÅ8*+jsKŠ*+Iž”)
:,Š*+:jsK8*+Iž”):,8*+;j +C{|,sK*+I¾Æ,+Ç<¼Æ,+
:[ÈÉI<=s*+,:;Ê`hb Â
G1H˓<=s*+, ,:;I¥¬Ì‡*+,:;Ã[h<= 10B 3 ,:;I(Ä( 10)
2 = 10
，32 = 9，10 > 9IÍ 10 > 3[
G2H%1., Âr/0+,ÎÏ)¹I·<=s+,:;[
G3HÐsKÑ!	,/0+<=:;ÒI‰F%hbÓÔ
① 𝑎> 𝑏≥0⇔𝑎>
𝑏[
② 𝑎> 𝑏⇔
3 𝑎>
3 𝑏[
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【培优篇】
【题型1  直接求平方根、立方根】
˜ 1G24-25 ÕÖºb·×ØÙÚ·ÛÜHbÝÞÂPßà,gG    H
A[ 5g 5 ,.4 !
B[0 , !A# !Žg 0
C[ 9, !g±3
D[−3g(−3)2,JK !
áâC
@ãäåæ !B.4 !,~IçèJ@ãéêë,ŠÌ9[
ì&&
êë A
 5g 5 ,.4 !I
.4 !g78+I­ 5íî( 5)
2
= 5IÍ A ŠÌj
êë B
0 , !A# !Žg 0I
 !A# !,‡ˆPI0 , !A# !ï‚ 0IÍ B ŠÌj
êë C
 9, !g ± 3I
9 = 3I¬ 3 , !F‚ ±
3I¬7 ± 3I ± 3, g 9IðÄ 9, !²ñßàIÍ C ßàj
êë D
−3g(−3)2,JK !I
(−3)2 = 9I9 , !‚ ± 3IòÈ−3g 9 ,JK !IÍ D ŠÌj
Íê C[
ó_ 1-1G24-25 ÕÖºb·ôõöØ÷·øùúûH−
8
27,# !g   j 4,.4 !g     [
áâ     
−
2
3     
2
@ãäåæ# !I.4 !I&,ügýþ
hÿJKŠ+𝑥, iÄ𝑎IĀāĂKŠ+𝑥
LM𝑎,.4 !jhÿJK+𝑦,# iÄ𝑏IĀāĂK+𝑦LM𝑏,# ![ăÈ&៉[
ì&&
−
8
27,# !g−
2
3j
∵4 = 2I2,.4 !g 2I
∴4,.4 !g 2[
Íáâ‚
−
2
3j 2[
ó_ 1-2G24-25 ÕÖºb·Ąą·ĆÛćĈHG1Hĉ{
a
0.000001
0.001
1
1000
1000000
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3 𝑎
G2H!ăĊċČ,č®ĉĎ
①
3 3 = 1.442I›
3 3000 = ______I
3 0.003 = ______j
②ďĐ
3 343 = 7I›
3 0.000343 = ______[
áâG1H0.01I0.1I1 I10 I100G2H①14.42I0.1442I ②0.07
@ãäđĒåæē# !,9–IvăQN +;+Ĕ¼ŒĔ¾ĕĖ 3 y”F,# !,;+
Ĕ¼ŒĔ¾ĕĖ 1 y&Ÿ‰IėúýþQN +;+B”F,# !;+ĕĖč®g&,ü[
G1H$%# !,9–&Ÿ‰j
G2H①$%# !,9–&Ÿ‰j
②$%# !,9–&Ÿ‰[
ì&&
G1H
3 0.000001 = 0.01j
3 0.001 = 0.1j
3 1 = 1j
3 1000 = 10j
3 1000000 = 100j
Íáâ‚
0.01I0.1I1 I10 I100j
G2H①
3 3000 =
3 1000 ×
3 3 = 10 × 1.442 = 14.42j
3 0.003 =
3 0.001 ×
3 3 = 0.1 × 1.442 = 0.1442j
Íáâ‚
14.42I0.1442j
②
3 0.000343 =
3 0.000001 ×
3 343 = 0.01 × 7 = 0.07
Íáâ‚
0.07[
ó_ 1-3™ a"b š‚f’+Ic"d š‚8œ+I› 𝑎2−𝑏2 +
3 𝑐𝑑=       
.
áâ―1
ì&!ăVĘ
a+b=0Icd=-1I
› 𝑎2−𝑏2 +
3 𝑐𝑑= (𝑎+ 𝑏)(𝑎−𝑏) +
3 −1 =
0−1=-1.
Íáâg:-1.
【题型2  利用求平方根、立方根解方程】
˜ 2G25-26 ęÖº·Ąą·Ě¹ćĈH”Ä*+𝑎,𝑏I‡ˆJd-.“ ⊕”
𝑎⊕𝑏= 𝑎2−2𝑏[Ä𝑥, 
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'(𝑥+ 1) ⊕ 8 = 0,&‚            
[
áâ𝑥1 = 3,𝑥2 = −5
@ã!ăV‡ˆ,ě-.IÝrÄ𝑥,JĜĝw 'Ÿ‰I&tŸ‰Ę½áâ.
ì&(V‰Ę(𝑥+ 1)2−16 = 0I
∴(𝑥+ 1)2 = 16,
∴𝑥+ 1 = 4Œ𝑥+ 1 = −4,
∴𝑥1 = 3,𝑥2 = −5[
Íáâ‚
𝑥1 = 3,𝑥2 = −5.
Ğäåæ,gJĜĝw '&,ğI*+,-.A0&‡ˆ,ě-.g&,ü.
ó_ 2-1G24-25 ÕÖºb·Ġ÷ġĢ·øùúûHbÝé_P, x[
(1)𝑥2 = 81j
(2)(𝑥+ 1)3 = −27[
áâ(1)𝑥=± 9j
(2)𝑥= −4j
@ãG1H!ă !,‡ˆ‘&𝑥,)IJKŠ+`sK !IģĤš‚f’+[
G2H!ă# !,‡ˆ‘&𝑥,)I¶r−27,# !I·Ęr𝑥+ 1,)I¥¬r𝑥[
äđĒåæē !A# !,‡ˆnF%[ėúýþ !A# !GhÿJK+,# iÄ𝑎IĀā
ĂK+LM𝑎,# !H,~g&,ü[
ì&G1H&
𝑥2 = 81I 
𝑥=±
81
𝑥=± 9j
G2H&
(𝑥+ 1)3 = −27I 
𝑥+ 1 =
3 −27I
𝑥+ 1 = −3I
∴x = −4[
ó_ 2-2G24-25 ÕÖºb·ĥ÷ĦħĨ·ÛÜHíî '(2𝑥−1)2−16 = 9, x ,)‚      
[
áâ3 Œ−2
@ãäåæē$% !& '[
¶>0Ę½(2𝑥−1)2 = 25I›2𝑥−1 = 5Œ2𝑥−1 = −5Iĩ.Ÿ‰[
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ì&&
∵(2𝑥−1)2−16 = 9
∴(2𝑥−1)2 = 25
∴2𝑥−1 = 5Œ2𝑥−1 = −5
∴𝑥= 3Œ𝑥= −2[
Íáâ‚
−2Œ 3[
ó_ 2-3G24-25 ÕÖºb·Īī·ÛPHbÝ 'P x ,)
(1)2𝑥2 = 50j
(2)
1
3(𝑥+ 1)2 = 3j
(3)(2−𝑥)3 = 64j
(4)64(𝑥+ 1)3−125 = 0
áâ(1)𝑥= 5Œ𝑥= −5
(2)𝑥= 2Œ𝑥= −4
(3)𝑥= −2
(4)𝑥=
1
4
@ãäđĒåæē$% !A# !ĬĐ+,)Iėúýþ&JK+, !n# !g&
,ü[
G1H!ă !,‡ˆ&Ÿ‰j
G2H!ă !,‡ˆ&Ÿ‰j
G3H$%# !,‡ˆ&Ÿ‰j
G4H$%# !,‡ˆ&Ÿ‰[
ì&G1H&
2𝑥2 = 50
𝑥2 = 25I
&Ę
𝑥= 5Œ𝑥= −5j
G2H&
1
3(𝑥+ 1)2 = 3
(𝑥+ 1)2 = 9I
𝑥+ 1 = 3Œ𝑥+ 1 = −3
&Ę
𝑥= 2Œ𝑥= −4j
G3H&
(2−𝑥)3 = 64
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2−𝑥= 4I
&Ę
𝑥= −2j
G4H&
64(𝑥+ 1)3−125 = 0
(𝑥+ 1)3 =
125
64 I
𝑥+ 1 =
5
4I
&Ę
x = 1
4[
【题型3  由平方根、立方根求值】
˜ 3G24-25 ÕÖºb·×ØĭĮ·øùúûHďĐ|𝑎−6|B 𝑎+ 2𝑏š‚f’+I𝑐+ 5,# !g 2I›
𝑎−2𝑏−𝑐, !‚     [
áâ ± 3
@ãÈđĒåæēf’+,‡ˆI# !I !IŠÌýþf‡ˆg&ü[¶$%# !"
š‚f’+,‡ˆĘr𝑎I𝑏I𝑐,)jįİ&Ęr𝑎−2𝑏−𝑐,)I·& !Ÿ‰[
ì&&
∵|𝑎−6|B 𝑎+ 2𝑏š‚f’+I
∴|𝑎−6| +
𝑎+ 2𝑏= 0I
∴𝑎−6 = 0I𝑎+ 2𝑏= 0I
&Ę
𝑎= 6I𝑏= −3I
∵𝑐+ 5,# !g 2I
∴𝑐+ 5 = 8I
∴𝑐= 3
∴𝑎−2𝑏−𝑐= 6−2 × (−3)−3 = 9
∴𝑎−2𝑏−𝑐, !g ± 3[
Íáâ‚
 ± 3[
ó_ 3-1ďĐ x ,sK !g𝑎+ 3B2𝑎−15I­2𝑏−1,.4 !g 3[
(1)𝑎,𝑏,𝑥,)j
(2)𝑎+ 𝑏−1,# ![
áâ(1)4I5I49
(2)2
@ãäåæ !I.4 !n# !,‡ˆ[ėúýþO‡ˆn9–g&,ü[
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G1H!ă !B.4 !,‡ˆŸ‰Ę𝑎I𝑏,)I·&𝑥,)Ÿ‰j
G2Hı𝑎I𝑏,)įİ𝑎+ 𝑏−1Pĩ.¹$%# !,‡ˆŸ‰Ęáâ[
ì&G1H&
 ∵𝑥,sK !g𝑎+ 3B2𝑎−15I­2𝑏−1,.4 !g 3I
∴𝑎+ 3 + 2𝑎−15 = 0I2𝑏−1 = 9I
&Ę
𝑎= 4I𝑏= 5j
∴𝑥= (𝑎+ 3)2 = 72 = 49j
G2H&
 ∵𝑎= 4I𝑏= 5I
∴𝑎+ 𝑏−1 = 4 + 5−1 = 8
ó_ 3-2G24-25 ÕÖºb·IJijĴ¼·øùúûHďĐ3𝑎+ 1,sK !@Sg7−𝑚A2−2𝑚I9 + 𝑏,
# !g 2[
(1)𝑚I𝑎I𝑏,)j
(2) 7𝑎−𝑏, ![
áâ(1)𝑚= 3I𝑎= 5I𝑏= −1
(2) ±
6
@ãÈåæ !I# !IėúýþfĐĵg&,üj
G1H!ă !‡ˆI# !‡ˆÝĘ7−𝑚+ 2−2𝑚= 0I9 + 𝑏= 23IŸ‰r𝑚I𝑎I𝑏,)j
G2H¶r7𝑎−𝑏,)I·$% !‡ˆrá⟉[
ì&G1H&
vVĘI7−𝑚+ 2−2𝑚= 0 
&Ę𝑚= 3        
Í7−𝑚= 4,2−2𝑚= −4
∴3𝑎+ 1 = 42I
&Ę𝑎= 5 
(V9 + 𝑏= 23I
&Ę𝑏= −1j
G2H∵7𝑎−𝑏= 7 × 5−(−1) = 36I
36 = 6 I6 , !‚ ±
6 I
Ķa 7𝑎−𝑏, !‚ ±
6[
ó_ 3-3G24-25 ÕÖºb·ķijěĸ·ÛÜHďĐ2𝑎+ 1, !g ± 3.3𝑎+ 2𝑏−4,# !g−2I
(1)𝑎A𝑏,)j
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(2)3𝑎+ 𝑏, !
áâ(1)𝑎= 4I𝑏= −8
(2) ± 2
@ãäđĒåæē# !I !,‡ˆI!ă# !I !,‡ˆ&Ÿ‰[
G1H(# !I !,‡ˆ‰Đ2𝑎+ 1 = 9I3𝑎+ 2𝑏−4 = −8I°¹Ÿ‰r aIb ,)[
G2HĹ aIb ,)įİ3𝑎+ 𝑏I· !Ÿ‰[
ì&G1H&
 ∵2𝑎+ 1, !g ± 3I3𝑎+ 2𝑏−4,# !g−2I
∴2𝑎+ 1 = 9I3𝑎+ 2𝑏−4 = −8I
∴𝑎= 4I
∴𝑏= −8[
G2H&
∵3𝑎+ 𝑏= 3 × 4−8 = 4I
∴3𝑎+ 2𝑏, !‚ ± 2[
【题型4  实数的运算】
˜ 4G24-25 ÕÖºb·ĺĻļ·ÛPHı 1I 2I 3I 6†hĽ _ľÝI™č‡(𝑚,𝑛){|Ŀ𝑚ľ»
¼Ĕ¾Ŀ𝑛K+I›(5,4)B(12,4){|,s+tŀg         
.
áâ0
@ãÈđĒåæē+Ł,ółč®I*+,-.Iх+Łółč®gü[
!ă+,ľÝ ‰ĐIĿJľ
1 K+IĿĝľ 2 K+[Ŀńľ 3 K+IĿľ 4 K+I…Ŀ𝑚ľ`𝑚K
+I»ĿJľ½𝑚ľŅ`
1 + 2 + 3 + 4 + … + 𝑚=
𝑚(𝑚+1)
2
K+I!ă+,ľÝ ÂIK+JKņŇI
!ăňVʼnŃrĿ𝑚ľĿ𝑛K+½ŊgŋK+¹·ĩ.[
ì&&
!ă+,ľÝ ‰ĐIĿJľ
1 K+IĿĝľ 2 K+[Ŀńľ 3 K+IĿľ 4 K+I…
Ŀ𝑚ľ`𝑚K+I»ĿJľ½𝑚ľŅ`
1 + 2 + 3 + 4 + … + 𝑚=
𝑚(𝑚+1)
2
K+I!ă+,ľÝ ÂIK
+JKņŇI
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(5,4){|Ŀ 5 ľ»¼Ĕ¾Ŀ 4 K+g 2I
∵Ō 11 ľŅ`1 + 2 + 3 + 4 + … + 11 =
11×(11+1)
2
= 66 (K)+I
∴ (12,4){|Ŀ 12 ľĿ 4 K+ŸĿ 70 K+I
∵70 ÷ 4 = 17⋯2I
∴ (12,4){|,+g 2I
∴ (5,4)B(12,4){|,s+tŀg 2−2 = 0I
Íáâ‚
0[
ó_ 4-1G2025 ōÖº·Ąą·ŎúûHĩ.
 16−(−1)2025−
3 27−|1−2|[
áâ3−2
@ãäđĒåæē*+,-.I¶ĩ..4 !A# !Iŏĩ.ž”)A© I¸¹ĩ.§¨
Ÿ‰Ę½áâ[
ì&&
Ő_ = 4−(−1)−3−( 2−1)
= 4 + 1−3−2 + 1
= 3−2[
ó_ 4-2G24-25 ōÖº·Ġ÷őļ·ÛÜHč‡
[𝑚]{|XŒ“𝑚,¸:>+I{𝑚}{|𝑚,;+?@I
[𝑚] + {𝑚} = 𝑚IOP𝑚‚*+[˜h[3.14] = 3I{3.14} = 0.14I[3.14] + {3.14} = 3.14[ĩ.
[−3]−{ 5}
=       
[
áâ−5
@ãäåæē쇈b,*+-.I/0+,1.I0&쇈g&,ü[@S1ĩ−3A 5
 ŋsK>+tuI·!ă쇈Ęr[−3] = −2I{ 5} =
5−2Isœf¨Ÿ‰Ęráâ[
ì&&
∵−2 < −3 < −1I2 <
5 < 3I
∴[−3] = −2I{ 5} =
5−2I
∴[−3]−{ 5} = −2−( 5−2) = −5[
Íáâ‚
−5[
ó_ 4-3G24-25 ÕÖº·Ŕķŕõ·ÛPHK'´ù(™ŖŗŘřŚ}IśŜŝĩēJù-.'´h
Ľ
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˜h
ÐŞİ x ,)‚−1ÒIĩ.Óÿ 3 < 4jıŞİ)ó‚(−1) +1ş0Iĩ.Óÿ‚ 6 < 4j·ıŞİ
)ó‚ē0 + 1 = 1IŠš-.I½ĩ.ÓÿX;Ä 4IŢŞrţÓÿ[
ŤʼnåbÝğ[
(1)ÐŞİ x ,)‚ 5I›Şr y ,)gxťŦŤÝ_ĩ.[
(2)ÐŧŨŞİ x ,)‚ 1IŤË“ĩ.ÞũŪ“cw'´-¦¹ŢzŞr y[
áâ(1) 21
(2)4 w
@ãäåæē*+,-.I0&VIýþūĽP,-.›g&,ü[
G1H!ăūĽP,-.'´ĩ.Ÿ‰j
G2H!ăūĽP,-.'´ĩ.I½Óÿ:ČiÄ 4 ŸŞrÓÿ‚Ŭ[
ì&G1HÐŞİ x ,)‚ 5 ÒI
›`5−(−2) = 7I7 × 3 = 21I
­ 21 > 4I
∴Şr y ,)g 21[
G2HÐŞİ x ,)‚ 1 ÒI
›`1−(−2) = 3I3 × 3 = 9I 9 < 4IŠšĩ.j
ĿĝwŞİ x ,)‚1 + 1 = 2ÒI
›`2−(−2) = 4I4 × 3 = 12I 12 < 4IŠšĩ.j
ĿńwŞİ x ,)‚2 + 1 = 3ÒI
›`3−(−2) = 5I5 × 3 = 15I 15 < 4IŠšĩ.j
ĿwŞİ x ,)‚3 + 1 = 4ÒI
›`4−(−2) = 6I6 × 3 = 18I 18 > 4IŞr𝑦j
ĶaŪ“ 4 w'´-¦¹ŢzŞr y[
【题型5  无理数的估算】
˜ 5G24-25 ÕÖºb·ŭŮůŰ·ÛÜH:ű°gŲ,ŝĩŜIŸųgJŴ;;,ŵŶIŷŸpŹ“Ų
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”[hĽI
𝐵𝐶
𝐴𝐵,)ÎźŻ<
5−1
2 I›bÝ1.ŠÌ,gG    H
A[0 <
5−1
2
<
2
5
B[
2
5 <
5−1
2
<
1
2
C[
1
2 <
5−1
2
< 1
D[1 <
5−1
2
< 2
áâC
@ãäåæē/0+,1.Iäåæ/0+,1ĩIXi_,9–IŠÌżŽ,¡¢g&ä
,ü[(VĐ4 < 5 < 9I›2 <
5 < 3I!ăXi_,9–ĩ.&Ÿ‰[
ì&&
(VĐI 4 < 5 < 9I
∴2 <
5 < 3I
∴1 <
5−1 < 2I
∴
1
2 <
5−1
2
< 1[
Íê
C[
ó_ 5-1G24-25 ōÖºb·ķžſƀ·øùúûHďĐ432 = 1849I442 = 1936I452 = 2025I462
= 2116[™𝑛‚>+­𝑛<
2022 < 𝑛+ 1I›𝑛,)‚G   H
A[43
B[44
C[45
D[46
áâB
@ãäđĒåæ/0+,1.Iėúýþ/0+,1.g&,üj¶!ăŖP,+ă1. 2022
,:;I¥¬ğ‰&[
ì&&
(V‰Đ
1936 < 2022 < 2025
∴44 <
2022 < 45I
∴𝑛= 44j
Íê B[
ó_ 5-2G24-25 ęÖº·ķžſƀ·ÛPH™𝑎<
23 < 𝑏I­𝑎"𝑏‚ƁšŠ>+I›𝑎+ 𝑏=        
áâ9
@ãäåæ*+,1.B:;<=,zƂI¶1.r 23,ƃ)¡¢IĘr𝑎I𝑏,)I¥¬‰ĘrÓ
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Ô[!ăVr𝑎I𝑏,)g&,ü[
ì&&
∵16 < 23 < 25I
∴4 <
23 < 5I
∵𝑎I𝑏‚sKƁš>+I
∴𝑎= 4I𝑏= 5I
∴𝑎+ 𝑏= 4 + 5 = 9[
Íáâ‚
9[
ó_ 5-3G24-25 ÕÖºb·ĥƄƅ×·ÛÜHďĐ𝑘<
17−1I­𝑘‚Š>+I›𝑘,)‰ag           
GqrJKŸ‰H[
áâ1GáâXƆJH
@ãäđĒåæē/0+,1."Xi_,&iĐĵIýþ/0+,1. Âg&,ü[
¶1. 17Ęƃ)¡¢I·Ì‡ k ,ƃ)¡¢I°¹!ăXi_&,‡ˆŸ‰&á[
ì&&
∵16 <
17 <
25g>+j
∴3 <
17−1 < 4gƇĄ +j
∴𝑘≤3I
∵𝑘‚Š>+I
∴𝑘= 1Œ 2 Œ 3[
Íáâ‚
1GáâXƆJH[
【拔尖篇】
【题型6  算术平方根的双重非负性】
˜ 6G24-25 ÕÖºb·ĠƄƈƉ·øùúûHďĐ aIb ‚*+I­𝑏=
𝑎−8−8−𝑎+25I›
3 𝑎+
𝑏,
)‚G    H[
A[10
B[9
C[8
D[7
áâD
@ãäåæ.4 !}#,ŗƊI# !I!ă.4 !,N +g78+Ę½𝑎= 8I¥¬
Ę𝑏= 25I¸¹įİ
3 𝑎+
𝑏ĩ.Ÿ‰[
ì&&
∵𝑏=
𝑎−8−8−𝑎+25I𝑎−8 ≥0I8−𝑎≥0I
∴𝑎= 8I𝑏= 25I
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∴
3 𝑎+
𝑏=
3 8 +
25 = 2 + 5 = 7I
Íê
D[
ó_ 6-1G25-26 ōÖº·Ąą·ƋƌúûH™𝑎+
𝑎−3 = 3I 𝑎+ 3,)[
áâ 6
@ãäđĒåæ,gĝw!_Iėúýþĝw!_,9–g&,ü[
¶ıŐ_ł‚ 𝑎−3 = 3−𝑎I·%Xi_Ś&Ÿ‰[
ì&&
ò‚𝑎+
𝑎−3 = 3I
Ķa 𝑎−3 = 3−𝑎I
Ķa𝑎−3 ≥0,3−𝑎≥0I
Ķa𝑎= 3I
Ķa 𝑎+ 3 =
3 + 3 =
6[
ó_ 6-2G24-25 ÕÖºb·ôõôļ·ÛPHďĐ|𝑎−3| +
3 + 𝑏= 0I›
𝑎
𝑏
2025
= G    H
A[1
B[−1
C[2
D[−2
áâB
@ãäå枔)I.4 !I`0+,© I&,ügr𝑎A𝑏,)[
!㞔)A.4 !,789I&Ę𝑎A𝑏,)Iįİĩ.Ÿ‰[
ì&&
|𝑎−3| +
3 + 𝑏= 0I|𝑎−3| ≥0I 3 + 𝑏≥0I
∴|𝑎−3| = 0I 3 + 𝑏= 0I
∴𝑎−3 = 0I3 + 𝑏= 0I
∴𝑎= 3I𝑏= −3I
∴
𝑎
𝑏= −1
∴
𝑎
𝑏
2025
= (−1)2025 = −1
Íê
B[
ó_ 6-3G2025 ōÖºb·ĠƄ·ŎúûHďĐ7«*+ aIb íî|2𝑎−4| + |𝑏+ 2| +
(𝑎−4)𝑏2
+4 = 2𝑎I›𝑎+ 𝑏=      
[
áâ2
@ãäđĒåæē.4 !,789Iž”),789Iı_ƍó^‚|2𝑎−4| + |𝑏+ 2| +
(𝑎−4)𝑏2
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= 2𝑎−4I(.4 !,789Iž”),789Ęr2𝑎−4 ≥0I·łƎ_ƍ‰Ęr|𝑏+ 2| +
(𝑎−4)𝑏2
= 0I·!ă.4 !,789Iž”),789Ÿ‰Ęr𝑏= −2I𝑎= 4I¥¬įİį+_Ÿ‰Ęrá
â[
ì&&
∵|2𝑎−4| + |𝑏+ 2| +
(𝑎−4)𝑏2 +4 = 2𝑎I
∴|2𝑎−4| + |𝑏+ 2| +
(𝑎−4)𝑏2 = 2𝑎−4I
∴2𝑎−4 ≥0I
∴2𝑎−4 + |𝑏+ 2| +
(𝑎−4)𝑏2 = 2𝑎−4I
∴|𝑏+ 2| +
(𝑎−4)𝑏2 = 0
∴𝑏+ 2 = 0I𝑎−4 = 0I
&ĘI𝑏= −2I𝑎= 4I
›𝑎+ 𝑏= 4−2 = 2I
Íáâ‚
2[
【题型7  实数的大小比较】
˜ 7G24-25 ÕÖºb·ĪīƏƐě·ÛPH<=:;
 11     3j−5     −2j
5+1
2     
3
2[Gĉ“ > ”"
“ = ”Œ“ < ”H[
áâ     
>     
<     
>
@ãäåæē*+,:;<=IėúýþaĐĵg&áä,ü[
!ă*+,:;<= Â&៉[
ì&&
 ∵ 
11 >
9I
∴
11 > 3j
∵5 >
4 = 2I
∴−5 < −4I
∴−5 < −2j
∵5 >
4 = 2I
∴
5 +1 > 3I
∴
5+1
2
>
3
2j
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Íáâ‚
 > I < I > [
ó_ 7-1G24-25 ÕÖºb·×ØƑõ·ÛPH™𝑎=
3 7I𝑏=
5I𝑐= 2I› aIbIc ,:;ÂG    H
A[𝑏≤𝑐< 𝑎
B[𝑏< 𝑎< 𝑐
C[𝑎< 𝑐< 𝑏
D[𝑎< 𝑏< 𝑐
áâC
@ãäåæ*+,:;<=I¶$%ƒƓÂ1. aIb ,)I·<=:;Ÿ‰[
ì&&
∵1 < 7 < 8I4 < 5 < 9I
∴
3 1 <
3 7 <
3 8I 4 <
5 <
9I
Ÿ1 <
3 7 < 2I2 <
5 < 3I
∴1 < 𝑎< 2I2 < 𝑏< 3I
Ɣ𝑐= 2I
∴𝑎< 𝑐< 𝑏I
Íê
C[
ó_ 7-2G25-26 ōÖº·Ąą·ƋƌúûH<= 65−2B 10 +2,:;[
áâ 65−2 >
10 +2
@ãÈåæ/0+,1.A*+,:;<=I‰Ë“1.sK/0+,¡¢I¥¬Ì‡sKį+_
,¡¢I·¥¦<=[
ì&&
ò‚8 <
65 < 9I3 <
10 < 4I
Ķa6 <
65−2 < 7I5 <
10 +2 < 6I
Ķa 65−2 >
10 +2[
ó_ 7-3G25-26 ōÖº·Ąą·ƋƌúûH<=bÝéŚ+,:;
(1)
1
2B
10
6 j
(2)
6−1
2 B 1[
áâ(1)
10
6 >
1
2
(2)
6−1
2
< 1
@ãäåæ*+<=:;Iėúýþ*+<=:;, ÂIg&,ü
G1H1.Â<=:;Ÿ‰j
G2H1.Â<=:;Ÿ‰[
ì&G1H&
∵10 >
9 = 3I
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∴
10
6 >
3
6IŸ
10
6 >
1
2j
G2H∵4 <
6 <
9I
∴2 <
6 < 3I
∴1 <
6−1 < 2I
∴
6−1
2
<
2
2IŸ
6−1
2
< 1[
【题型8  求整数部分和小数部分】
˜ 8G24-25 ōÖº·ĥ÷ěƕ·øùúûH™ 3,>+?@‚ aI;+?@‚ bI4−3,>+?@‚
cI;+?@‚ dI›
𝑏+𝑑
𝑎𝑐,)‚G   H
A[
1
2
B[
1
4
C[
3−1
2
D[
3+1
2
áâA
@ãäåæē1./0+,:;I˓1. 3 ŋsK>+tuI»¬Ì‡𝑎= 1I𝑏=
3−1IË
“1.4−3 ŋsK>+tuI»¬Ì‡𝑐= 2I𝑑= 2−3I°¹Ĺ a"b"c"d įİĘ+)[
ì&&
∵3,>+?@‚ aI;+?@‚ bI1 <
3 < 2I
∴𝑎= 1I𝑏=
3−1I
∵1 <
3 < 2I
∴−2 < −3 < −1I
∴2 < 4−3 < 3I
∵4−3,>+?@‚ cI;+?@‚ dI
∴𝑐= 2I𝑑= 4−3−2 = 2−3I
∴
𝑏+𝑑
𝑎𝑐=
3−1+2−3
1×2
=
1
2[
Íê
A[
ó_ 8-1G24-25 ÕÖºb·IJØIJõ·ÛPH™6 +
5,>+?@g𝑚I;+?@g𝑛I›|𝑛−𝑚|‚G   H
A[ 5−10
B[10−5
C[ 5−6
D[6−5
áâB
@ãäåæB/0+,>+?@`,ĩ.I*+,-.IƒƓÂr𝑚,𝑛,)I·į)ĩ.Ÿ‰[
ì&&
∵4 <
5 <
9I
∴2 <
5 < 3I
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∴8 < 6 +
5 < 9I
∴𝑚= 8I𝑛= 6 +
5−8 =
5−2I
∴|𝑛−𝑚| = | 5−2−8| = 10−5j
Íê B[
ó_ 8-2G24-25 ōÖº·Ġ÷őļ·ÛÜHč‡
[𝑚]{|XŒ“𝑚,¸:>+I{𝑚}{|𝑚,;+?@I
[𝑚] + {𝑚} = 𝑚IOP𝑚‚*+[˜h[3.14] = 3I{3.14} = 0.14I[3.14] + {3.14} = 3.14[ĩ.
[−3]−{ 5}
=       
[
áâ−5
@ãäåæē쇈b,*+-.I/0+,1.I0&쇈g&,ü[@S1ĩ−3A 5
 ŋsK>+tuI·!ă쇈Ęr[−3] = −2I{ 5} =
5−2Isœf¨Ÿ‰Ęráâ[
ì&&
∵−2 < −3 < −1I2 <
5 < 3I
∴[−3] = −2I{ 5} =
5−2I
∴[−3]−{ 5} = −2−( 5−2) = −5[
Íáâ‚
−5[
ó_ 8-3G24-25 ÕÖºb·IJØķƖ·øùúûHďĐ*+8 +
13,>+?@‚𝑎I;+?@g𝑚j*+
9−13,>+?@‚𝑏I;+?@g𝑛[
(1)qr𝑎I𝑚I𝑏I𝑛,)j
(2)12𝑚+ 12𝑛+
𝑎+ 𝑏−7,), !j
(3)
3 (𝑚+ 𝑛)2025 +(𝑎−2𝑏)+ 6,)[
áâ(1)𝑎= 11I𝑚=
13−3I𝑏= 5I𝑛= 4−13
(2) ± 3
(3)2
@ãäåæē/0+,1." !"# !IėúýþaĐĵŏƗƘ-%g&È,ü[
G1H1.Ęr3 <
13 < 4I»¬‰Ę11 < 8 +
13 < 12I5 < 9−13 < 6IÓEVŸ‰Ę&j
G2HıG1HP𝑎= 11I𝑚=
13−3I𝑏= 5I𝑛= 4−13įİĶ_ƍ¥¦ĩ.I·ÓE !,‡ˆĩ
.Ÿ‰Ę&j
G3HıG1HP𝑎= 11I𝑚=
13−3I𝑏= 5I𝑛= 4−13įİĶ_ƍ¥¦ĩ.I·ÓE# !,‡ˆĩ
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.Ÿ‰Ę&[
ì&G1H&
∵9 < 13 < 16I
∴9 <
13 <
16IŸ3 <
13 < 4I
∴11 < 8 +
13 < 12I5 < 9−13 < 6I
∵*+8 +
13,>+?@‚𝑎I;+?@g𝑚I*+9−13,>+?@‚𝑏I;+?@g𝑛I
∴𝑎= 11I𝑚= 8 +
13−11 =
13−3I𝑏= 5I𝑛= 9−13−5 = 4−13j
G2H&
(G1H‰Ę
𝑎= 11I𝑚=
13−3I𝑏= 5I𝑛= 4−13I
12𝑚+ 12𝑛+
𝑎+ 𝑏−7
= 12 × ( 13−3) + 12 × (4−13) +
11 + 5−7
= 12 13−36 + 48−12 13 + 4−7
= 9I
∴12𝑚+ 12𝑛+
𝑎+ 𝑏−7,), !‚ ±
9 =± 3j
G3H&
3 (𝑚+ 𝑛)2025 +(𝑎−2𝑏)+ 6
=
3 ( 13−3 + 4−13)2025 + (11−2 × 5) + 6
=
3 1 + 1 + 6
=
3 8
= 2[
【题型9  实数与数轴综合运用】
˜ 9G24-25 ÕÖºb·ĥƄƙƚƛ·ÛÜHƜęƝ.4ƞPƟƠ4ơ
“ƟƠéű©Iŏ¬N ªtIŸ
Ƣ”IŸ𝑐=
𝑎2 + 𝑏2G𝑎‚“Ɵ”I𝑏‚“Ơ”I𝑐‚“Ƣ”H™“Ɵ”‚2I“Ơ”‚3I›“Ƣ” hĽĶ|+C‰{|
 G    
H
A[𝐴
B[𝐵
C[𝐶
D[𝐷
áâC
@ãäåæ/0+,1.I*+B+CI0&VŏÝʊÌ,._g&,ü[!ăVÝ_
ĩ.¹1.O:;I°¹Ì‡O +C,yƣŸ‰[
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ì&&
™“Ɵ”‚2I“Ơ”‚3I› 22 + 32 =
13I
∵9 < 13 < 16I
∴3 <
13 < 4I
›“Ƣ” hĽĶ|+C‰{| 𝐶I
Íê
C[
ó_ 9-1G24-25 ÕÖºb·ƤÚƥ×·ÛPHhĽIŠ ^𝐴𝐵𝐶𝐷,ÁƦ‚ 7IƧ A  +C{|,+
‚−1I™ E  +CG E   A ,¼ƨHI­𝐴𝐷= 𝐴𝐸I› E Ķ{|,+‚          
[
áâ−7−1
@ãäåæ.4 !I*+B+CI¶rŠ ^,ÆġI¥¬!ăsu,Ʃƪr E Ķ{
|,+Ÿ‰[
ì&&
∵Š ^𝐴𝐵𝐶𝐷,ÁƦ‚ 7I
∴𝐴𝐸= 𝐴𝐷=
7I
∵Ƨ A  +C{|,+‚−1I
∴ E Ķ{|,+‚−7−1j
Íáâ‚
−7−1[
ó_ 9-2G24-25 ōÖºb·ƫůķƬƭķ·ÛÜH*+𝑎I𝑏 +C,yƣhĽĶ|IłƎ
|𝑎+ 1|−(𝑏−1)2
+
(𝑎−𝑏)2 =           
[
  
áâ2
@ãäåæē*+B+CI$%+CĘr−1 < 𝑎< 0I1 < 𝑏< 2I¥¬łƎŸ‰[
ì&&
(+CIĘ−1 < 𝑎< 0I1 < 𝑏< 2I
∴𝑎+ 1 > 0I𝑏−1 > 0I𝑎−𝑏< 0I
∴Ő_ = 𝑎+ 1−|𝑏−1| + |𝑎−𝑏|
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= 𝑎+ 1−(𝑏−1)−(𝑎−𝑏)
= 𝑎+ 1−𝑏+ 1−𝑎+ 𝑏
= 2I
Íáâ‚
2[
ó_ 9-3G24-25 ÕÖºb·Ġ÷\Ƒ·ÛPHhĽI˓ƮÆġ‚ 1 ,Š ^IƯz…Ì,Ĺ 2{| +
C𝐴1ưIƱ𝐴1¾ƨ¸Î,>+‚𝐵1Ia𝐵1‚kƲI𝐴1𝐵1‚ƳƴƮƳkIƵ+CĐ𝐴2IƱ𝐴2¾ƨ
¸Î,>+‚𝐵2Ia𝐵2‚kƲI𝐴2𝐵2‚ƳƴƮƳkIƵ+CĐ𝐴3IhÈŠšI›𝐴2026𝐵2026,ġ
‚          
[
áâ 2−1/−1 +
2
@ãäåæē*+,-.,č®I*+B+CI¶r B {|,+Ę½𝐴1𝐵1 = 2−2I›𝐴2{|
,+‚4−2I·r𝐵2{|,+‚3I›𝐴2𝐵2 =
2−1I°¹vw{|𝐴3𝐵3 = 2−2I𝐴4𝐵4 =
2−1I𝐴5𝐵5
= 2−2j𝐴6𝐵6 =
2−1jŸ‰Ń½č®&[
ì&&
(VĘI𝐴1{|,+‚ 2I
∵1 <
2 <
4I
∴1 <
2 < 2I
∴𝐵1{|,+‚ 2I
I∴𝐴1𝐵1 = 2−2I›𝐴2{|,+‚2 + 2−2 = 4−2I
∵1 <
2 < 2I
∴−2 < −2 < −1
∴2 < 4−2 < 3I
∴𝐵2{|,+‚3I
∴𝐴2𝐵2 = 3−(4−2) =
2−1I
µ0‰Ę𝐴3𝐵3 = 2−2j
𝐴4𝐵4 =
2−1j
𝐴5𝐵5 = 2−2j
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𝐴6𝐵6 =
2−1j
……I
aȃƶ‰ĘIÐ n ‚Ʒ+Ò𝐴𝑛𝐵𝑛= 2−2IÐ n ‚Ƹ+Ò𝐴𝑛𝐵𝑛=
2−1I
∴𝐴2026𝐵2026 =
2−1
Íáâ‚
 2−1[
【题型10  实数运算的应用】
˜ 10G24-25 ÕÖºb·Ġ÷ġĢ·øùúûHƹŜƺ`JKƻƦ‚200cm3,ƼƽŐƾIƹŜƺƿĒıĂ
KƼƽǀłŏ6ěǁǂ}ě,^ǃ[
(1)™ıŐƾ6ěǁǂ}JKŊÁ‚Š ^"DŽ‚8cm,ġ ƻIġ ƻŊÁŠ ^,Æġ[
(2)ƹŜƺČıŐƾǁǂ}ńK:;fµ,Š ƻƼƽIćƇ}¹Džb,ĸƾƻƦ‚8cm3Ić},
JK;Š ƻƼƽ,džġ[
áâ(1)5cm
(2)4cm
@ãäåæ.4 !I# !,F%I
G1H!ăġ ƻƻƦ,ĩ._“ġ ƻ,ƻƦ = ŊÁƦ × DŽ”Ý '&Ÿ‰j
G2H!ă“Š ƻƻƦ,ĩ. Âan3K;Š ƻƻƦBÇƻƦtu,Ã”Ý '&Ÿ‰j
0&.4 !"# !,‡ˆgŠÌ&á,ü[
ì&G1H&
ŝġ ƻŊÁŠ ^,Æġ‚𝑥cmI
vVIĘ
8𝑥2 = 200I
&Ę
𝑥= 5Œ𝑥= −5G8)XTEVILJLjHI
á
ġ ƻŊÁŠ ^,Æġ‚5cmj
G2H&
ŝJK;Š ƻƼƽ,džġ‚𝑦cmI
vVIĘ
3𝑦3 +8 = 200I
&Ę
𝑦= 4I
á
JK;Š ƻƼƽ,džġ‚4cm[
ó_ 10-1G24-25 ÕÖº·Ŕķ·ÛPHhĽIJKŊÁƳƴ‚3cm,ljƍ£NJŹJoNjnj[ÐljƍŠ
ǍÒIlj£Njnj,DŽǎ‚16cmjœǍÒIĎĸ?@,DŽǎ‚4cm[lj£,NjnjŠǏœí 2 KJǐ:,Š 
ƻǑǒGπƃ 3IǑǒ,ǓǎXĩH[
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(1)ţljƍ,ǑƦGNJíÒNjnj,ƻƦHgxť# ǔǕŦ
(2)Š ƻǑǒ,džġgxťǔǕŦ
áâ(1)540cm3
(2)6cm
@ãäåæ`0+,²E-."JK+,# !IÊǖn\]cƻ,ƻƦIǗǘVIĘr“lj
ƍ,ǑƦBµŊ"DŽ‚(16 + 4)cm,kǙƻƦfi”g&,ü[
G1Hljƍ,ǑƦBµŊ"DŽ‚(16 + 4)cm,kǙƻƦfiI(ȉ&j
G2Hǚ¶rlj£,Njnj,ƻƦI°¹!ălj£,NjnjŠǏœí 2 KJǐ:,Š ƻǑǒ&Ÿ‰[
ì&G1H&
𝑉= π𝑟2ℎ≈3 × 9 × (16 + 4) = 27 × 20 = 540(cm3)j
G2H&
ò‚𝑉溶液= π𝑟2ℎ≈3 × 32 × 16 = 432(cm3)[
Ķadžġ =
3 432 ÷ 2 = 6cm[
ó_ 10-2JǛǜĚ¹I‚ē{ǝ ĆÛvǞǟǠƇ}+ćĈ,;ǡA;ǢµI+śŜǣ‡
 ǤÉǥǦ 2 ǧǨǩŽg 16 Ĝ,ǪǫǬǭǮĤ[»ǪǫĨ½ǜ,ǯǰǨDZDzdzŽ‚ 1.6 ĜI(
Ä+śŜgţǦ,ǴǵIòȍǨŽ‰aų%JKǦǶDz, 5 ĜǦǷǸ”ǯǰǨ,Çǩ¨ǹ 5
ĜGǰǨÇǩXpDZDzdzIµJǰǨǺǻǼų%JKǷǸH[ǽ!ăţǪǫĨ,ǾƘĖIÐǰǨÇǩGX
pDZDzdzHí 30 ĜÒI5 Ĝ,ǦǷǸ‰ǿȀ‚JK 7 Ĝ,ĨȁǷǸIŸ‰aȂǰǨÇǩGXpDZDzdzH
¨ǹ 7 ĜÐ+śŜµÒē 2 ǧǪǫ…ȃbǨȄȅÒI;ǡµÞ
“śŜIȆĤ‰aȀJdbǨ _I
ǾȇxȈ”ŤµĤ@ã;ǡµ,bǨ _Iŏĩ.räwÉǥÇdz%GǸpDZDzdzH¸ȉ‰‚        
Ĝ[
áâ25.2
@ã@Sĩ.sdbǨ, _I<=ŋJdÇdz%ȇȉŸ‰[
ì&ĿJdbǨ _‚ȊȋsǧǪǫ
Eĩdz%‚
16 + 16 + 1.6−7 = 26.6Ĝ
ĿĝdbǨ _‚bsKǰǨIKǰǨȋJǧǪǫ
Eĩdz%‚
(16 + 1.6−5) × 2 = 25.2Ĝ
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ÍêȌĿĝdȇȍ.I¸ȉdz%‚ 25.2 Ĝ
Íáâ‚
25.2[
Ğäåæē*+-.,*ȎF%I@ƒȏÔg&,ü[
ó_ 10-3Ǥ;R…ȃNċJƽġ‚32mIȐ‚21m,ġ ^ĎȑI
(1) âJ
hĽIıĂƽĎȑdȒȓIPuȔJŗȕȖ,;ȗI;ȗ,¼ÆȘĔ¾ĕ1mƯgģ,¾Æ
Ș[›ĂƽȒȑ,ÁƦ‚ _____ m2j
(2) âĝ
ȔŮJKġgȐ, 1.5 șIÁƦ‚384m2,ȚțȜI™<ȝ%,ȚțȜĒġ 23m½30mt
uIȐ 13m½20mtu[ ĂKȚțȜz%M<ȝȞŦ ŏÞũ0([
áâ(1)651
(2)zI0(]&ã
@ãäåæēĽ^,ĕI !,‡ˆiĐĵ[
G1H(VIȒȑ,ġ¨;1mIȐXóIò¬‰ĘȒȑ,ÁƦj
G2HŝȐ𝑥mI›ġ‚1.5𝑥mI!ăÁƦ_Ÿ‰ĘÄ x , 'I( !,‡ˆŸ‰Ę xI·” x ,
)¥¦1.I™íîVŸ‰Iȟ›X¦[
ì&G1H&
(VI;ȗ,¼ÆȘĔ¾ĕ1mƯgģ,¾ÆȘŸ;ȗ,Ȑ‚1mI
›Ȓȑ,ġ¨;1mIȐXóI
ÁƦ‚21 × (32−1) = 651m2j
Íáâ‚
651[
G2HzI0(hb
ŝȐ𝑥mI›ġ‚1.5𝑥mI
vV`
1.5𝑥2 = 384I 𝑥2 = 256
∵𝑥> 0I
∴𝑥= 16I1.5𝑥= 24
TEġ 23m½30mtuIȐ 13m½20mtuI
∴ĂKȚțȜz%M<ȝ[
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