初三数学知识手册（上）-答案_2026秋初中教辅合集_2027《新东方•知识手册》9上数学

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1."#$%
二次方程!"#$%&
C 
"&'%()*+,-./0123456 
78-9:;<=:;<>?0@A 2 03BC12D-./01
26 
EF A GHI 𝑎 JKA 0 L8EMNOPQ-4J-./012 
 
2."#$%C 
"&'%RSCTUV𝑥(𝑥+ 3) −(𝑥+ 3) = 0N(𝑥−1)(𝑥+ 3) = 0N 
TUV𝑥! = 1N𝑥" = −36 
WG C 
 
3."#$%B 
"&'%X𝑘= 0YN−𝑥+ 4 = 0NZY𝑥= 4N8[<\] 
X𝑘≠0YN∵12𝑘𝑥" −𝑥+ 4 = 08[<\N 
∴^= (−1)" −4 × 𝑘× 4 ⩾0N 
&UV𝑘⩽
!
!#N 
ZY𝑘⩽
!
!#=𝑘≠0] 
_`N𝑘⩽
!
!#6 
WG B 
 
4."#$%D 
"&'%2014 abcdefg 2500 h. 
2015 abcdefg 2500(1 + 𝑥) h. 
2016 abcdefg 2500(1 + 𝑥)" h. 
iajkbc 1.2 l.Nm2500 + 2500(1 + 𝑥) + 2500(1 + 𝑥)" =
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120006 
WG D 
 
5."#$%D 
"&'%∵ 3𝑥" + 𝑎𝑥+ 𝑏= 0 3n9\A −2 o 4  
∴ 3𝑥" −𝑎𝑥+ 𝑏= 3(−𝑥)" + 𝑎(−𝑥) + 𝑏= 0 IN−𝑥! = −2N−𝑥" =
4 
∴ 𝑥! = 2N𝑥" = −4Npqr&3123n9\A 2 o −4  
∴ 3𝑥" −𝑎𝑥+ 𝑏= 0 Tr&A 3(𝑥−2)(𝑥+ 4) = 0 
WG D 
 
6."#$%A 
"&'%s 𝑦= 𝑥" −2𝑥NmS12tAV𝑦−
!"
$ = −1 
BuUV(𝑦−3)(𝑦+ 4) = 0 
&UV𝑦= 3 v 𝑦= −4 
X 𝑦= 3 YN𝑥" −2𝑥= 3 
BuUV(𝑥−3)(𝑥+ 1) = 0 
&UV𝑥! = 3N𝑥" = −1 
X 𝑦= −4 YN𝑥" −2𝑥= −4 
∵w = −12 < 0 
∴ Z12x& 
fyzN𝑥" −2𝑥 3{A 3 
WG A 
 
7."#$%
%
&]
!
!# 
"&'%S12TP|AV𝑥" +
%
" 𝑥+
!
" = 0 
}HUV𝑥" +
%
" 𝑥= −
!
" 
!
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~1UV𝑥" +
%
" 𝑥+ (
%
&)" = −
!
" + (
%
&)" 
BuUV(𝑥+
%
&)" =
!
!# 
•€•‚ƒ<UV{
ℎ=
%
&
𝑘=
!
!#
 
 
8."#$%0 
"&'%„𝑥= 𝑎…c12TUN 
𝑎" −𝑎−1 = 0Np𝑎" = 𝑎+ 1N 
∴𝑎& −3𝑎−2 = (𝑎")" −3𝑎−2 
= (𝑎+ 1)" −3𝑎−2 
= 𝑎" −𝑎−1 = 06 
 
9."#$%4 
"&'%\†‡ˆ4uUV𝑥! + 𝑥" = −
'
( = 6 
‰𝑥! = 2 
Š𝑥" = 46 
 
10."#$%12 v 16 
"&'%∵ 12 𝑥" −8𝑥+ 𝑚= 0 8n9\ 
∴w = (−8)" −4𝑚≥0N&UV𝑚≤16 
R\‹ƒ<3ŒƒTUV𝐴𝐵+ 𝐴𝐶= 8N𝐴𝐵⋅𝐴𝐶= 𝑚 
∵ •Ž △𝐴𝐵𝐶 3-• 𝐵𝐶 3•A 6 
∴𝐴𝐵N𝐴𝐶 3•r‘J 4’4 v 2’6 v 6’2 
X 𝐴𝐵N𝐴𝐶 3•r‘J 4’4 YNp12 𝑥" −8𝑥+ 𝑚= 0 8n
9“•3[\ 
ZY w = (−8)" −4𝑚= 0N&UV𝑚= 16 
𝐴𝐵N𝐴𝐶 3•r‘J 2’6 v 6’2 YNp12 𝑥" −8𝑥+ 𝑚= 0 
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8n9Q“•3[\ 
ZY w = (−8)" −4𝑚> 0N𝐴𝐵⋅𝐴𝐶= 2 × 6 = 𝑚N&UV𝑚= 12 
∴𝑚 3{A 12 v 16 
 
11. 
(1) "#$%𝑥! = 0N𝑥" = 2 
"&'%𝑥" −2𝑥= 0N 
𝑥(𝑥−2) = 0N 
𝑥= 0v𝑥−2 = 0N 
OP𝑥! = 0N𝑥" = 2] 
 
(2) "#$%𝑥! =
%)√!%
"
N𝑥" =
%+√!%
"
 
"&'%^= (−3)" −4 × (−1) = 13N 
𝑥=
%±√!%
"×! N 
OP𝑥! =
%)√!%
"
N𝑥" =
%+√!%
"
] 
 
(3) "#$%𝑥! = 𝑥" = √5 
"&'%𝑥" −2√5𝑥+ 5 = 0N 
(𝑥−√5)" = 0N 
OP𝑥! = 𝑥" = √56 
 
12. 
(1) "#$%(20 + 2𝑥);(40 −𝑥) 
"&'%s”•–—˜™𝑥.YN”šT›œ(20 + 2𝑥)•N”••ž(40 −
𝑥).NW#$AV(20 + 2𝑥)N(40 −𝑥)] 
(2) "#$%˜™20.v10. 
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"&'%\†)FNUV(20 + 2𝑥)(40 −𝑥) = 1200 
&UV𝑥! = 20N𝑥" = 10 
#V”•–—˜™20.v10.NŸ ”š¡ž1200.] 
 
(3) "#$%Q¢ 
"&'%Q¢N 
∵(20 + 2𝑥)(40 −𝑥) = 2000  Z12x&N 
WQT¢£¤Ÿ ”š•ž2000.6 
 
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1."#$%
二次函数!"#$%&
B 
"&'%A (𝑎)*0+,-./012, 
B 34/01256, 
C 7 𝑦= 2𝑥+ 1.8012, 
D 9:07;0,-34/01256, 
<= B> 
 
2."#$%D 
"&'%
 
<= D 
 
3."#$%D 
"&'%?@A𝑥BC,DE?@FGHI.0>JKL#> 
MJNO,P
4𝑐−(−8)2
4×1
= 0, 
&P𝑐= 16> 
<= D 
 
4."#$%B 
"&'%Q𝑦= 0,PRST𝑘𝑥) −6𝑥−9 = 0,UVW-)X2M,! > 0> 
<= B> 
 
5."#$%C 
"&'%Q𝑥= 0,𝑦= 1,Y 𝐶 @FHI7(0"1)Z 
Q𝑦= 0,𝑥= ±1,Y𝐴(−1,0),𝐵(1"0)> 
A>[𝐴#𝐵#𝐶;@HI\EP]𝐴𝐶= 𝐵𝐶,^𝐴𝐶) + 𝐵𝐶) = 𝐴𝐵) 
Y△𝐴𝐵𝐶.)_`a;ab,<c=defZ 
(
)
2
2
2
4
1
4
4 3
2
3
y
x
x
x
x
x
=
-
+ =
-
+ - =
-
-
!
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6."#$%
B>g/012𝑦= −𝑥) + 1Fh2d71,
i/012𝑦= −𝑥) + 1Fjkl y Bm*@𝐶(0"1),
n@𝐶FHI.(0"1),<c=defZ
C>/012𝑦= −𝑥) + 1FjkFopB.𝑦B,n𝑥= 0Z<c=d 
qrZ
D>(𝑥$0+,𝑦s𝑥tuvwxZ<c=defZ
<=yC>
D 
"&'%[jz\Py 
𝑎< 0,< A qrZ 
(𝑥> 1+,𝑦s𝑥Ftuvwx,< B qrZ 
𝑐> 0,< C qrZ 
[op{\|,−1"3.ST𝑎𝑥) + 𝑏𝑥+ 𝑐= 0 FVWM,< D ef> 
 
7."#$%C 
"&'%[8012jk\|PR𝑏< 0,𝑎> 0,/012&'}~/0d•
2u*0,80d•2x*0,€K/012jk•‚ƒC,opB
A𝑦B„…> 
 
8. 
(1) "#$%1 
"&'%∵ [†~𝑥‡𝑦Foˆ‰\|,(𝑥= −1l𝑥= 3+𝑦F‰Š), 
∴ opB.`‹𝑥=
*+,-
)
= 1> 
<#$71 
(2) "#$%𝑦= −𝑥) + 2𝑥+ 8 
"&'%Œ•‹F&'}7y𝑦= −𝑥) + 2𝑥+ 8> 
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(3) "#$%𝑘 FŽ‰••.𝑘⩽9 
"&'%∵ [(2)|Œ•‹F&'}7𝑦= −𝑥) + 2𝑥+ 8, 
∴ ‘?@HI7y(1,9), 
∴ (ST𝑎𝑥) + 𝑏𝑥+ 𝑐= 𝑘U&+𝑘FŽ‰••.𝑘⩽9> 
 
9. 
(1) "#$%𝑦= (𝑥−3)) −1Z 
"&'%𝑦= 𝑥) −6𝑥+ 8 = (𝑥−3)) −1Z 
(2) "#$%−1,8 
"&'%’j\Py(0 ⩽𝑥⩽4+,𝑦F9x‰.−1,9u‰.8> 
(3) "#$%2 < 𝑥< 4> 
"&'%Q 𝑦= 0,𝑥2 −6𝑥+ 8 = (𝑥−2)(𝑥−4) = 0, 
∴𝑥+ = 2, 𝑥) = 4, 
∵ Œ•‹•‚ƒC, 
∴𝑦< 0 +,2 < 𝑥< 4> 
 
10."#$%𝑦= −2(𝑥−3)) −2 
"&'%“”•FŒ•‹b–-—,?@7𝑃(3, −2)F˜Œ•‹F†™}7 
𝑦= −2(𝑥−3)) −2> 
 
11."#$%−4 < 𝑥< −3 
"&'%∵@𝐴FšHI7−4,@𝐴›@𝐵œ*Œ•‹FopBop,@𝐵Fš
HI71, 
∴Œ•‹FopB7y𝑥= −
-
), 
∵/012𝑦1 = 𝑎𝑥2 + 𝑏𝑥+ 𝑐l8012𝑦2 = 𝑘𝑥Fjkm*@𝐴›•@
𝑂, 
∴/012l𝑥Bž8m@ 𝐶 HI7(−3,0), 
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YŸ 0 < 𝑦+ < 𝑦)F𝑥FŽ‰••.y−4 < 𝑥< −3> 
 
 
 
12. 
(1) "#$%𝑦= −
+
- 𝑥) +
/0
- 𝑥Z(0 < 𝑥< 46) 
"&'%[NOPy𝑦=
/0*1
-
𝑥= −
+
- 𝑥) +
/0
- 𝑥, 
∵
/0*1
-
> 0, 
∴𝑥< 46, 
<y𝑦= −
+
- 𝑥) +
/0
- 𝑥,(0 < 𝑥< 46)Z 
(2) "#$%¡ ( 𝑥= 24 +,𝑦 9u‰ 192 “S¢Z£ 2 ¢> 
"&'%¡[NOPy𝑦=
/2*1
-
⋅𝑥= −
+
- (𝑥−24)) + 192, 
<y(𝑥= 24+,𝑦[9u‰ 192 “S¢Z 
£[NOPy
48−𝑥
3
⩽10,&Py𝑥⩾18, 
∴(𝑥= 24¢+,
48−𝑥
3
= 8%, 
<y¤¥¦F§‚lx¨F©ª710 −8 = 2¢> 
 
13. 
(1) "#$%𝑦= −4𝑥) + 380𝑥−7800 
"&'%«¬2-®𝑚¯†°, MJNOP, «¬2-𝑚›«¬±²𝑥³80
12ϥ 
´8012œ•}7𝑚= 𝑘𝑥+ 𝑏 
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{50𝑘+ 𝑏= 60
49𝑘+ 𝑏= 64 &P{𝑏= 260
𝑘= −4 
i«¬2-𝑚›«¬±²𝑥F12œ•7𝑚= −4𝑥+ 260 
i𝑦= 𝑚(𝑥−30) = (−47 + 2060)(𝑥−30) 
µ¶F«¬·¸𝑦l«¬±²F12†™}7𝑦= −4𝑥) + 380𝑥−
7800 
 
(2) "#$%𝑥= 47¹48+,DE𝑦F9u‰.1224 
"&'%𝑦= −4𝑥) + 380𝑥−7800 = −4(𝑥−
56
) )) + 1225 
Œ•‹F•‚ƒº, opB.𝑥=
56
) , 
€7±²7»2,DE𝑥= 47¹48+ 
DE𝑦F9u‰.1224 
(3) "#$%ˆ¼½A45¾¿55¾À© 
"&'%(𝑦= 1000+, −4(𝑥−
56
) )) + 1225 = 1000 
&ÁWST, P𝑥1 = 40, 𝑥2 = 55 
(40 ⩽𝑥⩽55+, µ¶F«¬·¸-Â* 1000 ¾ 
[µ¶Fóc-ÄÅ 2400 ¾, P30(−4𝑥+ 260) ⩽2400, 
&ÁW-)}P, 𝑥⩾45 
∴45 ⩽𝑥⩽55 
«¬±²ˆ¼½A45¾¿55¾À©> 
 
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1. "#$%
旋转!"#$%&
B 
"&'%A()*+,-./012345)6789: 
B(*+,-./0123456789: 
C()*+,-./012345)6789: 
D()*+,-./012345)6789: 
24 B 
 
2. "#$%D 
"&'%D 45;</=>?-.@A124 D 
 
3."#$%C 
"&'%;BCDEF@1GHIJ+D/0KL 𝑶 MNOBC 𝟗𝟎! P@
AD/0* C +D/0 
24 C 
 
4."#$%B 
"&'%AQRS>T+DUVDRS1)*WKXYZELDCS12)
*BC: 
BQ[\D]O^SD>T1*BC_`: 
CQaVbcD^S1*de_`: 
DQXY/0fghi-jD>T1*?-.: 
24 B 
 
5."#$%𝟐 
!
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"&'%klm@1/0noYpqrstu 
v 2009 ÷ 3 = 669 w 2  
x3qryzD{ 2009 Y/$*{ 2 Y 
 
6."#$%(4) 
"&'%kl/01G/0 (1) |L 𝑂 }BC+,1n~•NOBC 
90∘1@A€XY/01nBC•~‚A<ƒD„… 
v 2015 = 503 × 4 + 3 
† { 2015 ~BCPD/0‡ (4) Xˆ 
 
7."#$%(5,2) 
"&'%v L 𝑀(𝑥, 𝑦) KL 𝑂 •NOBC 90∘ @A 𝑀!(𝑦, −𝑥) 
† -‰L 𝐴! DŠ‹* (5,2) 
 
8."#$% 90  
"&'%v ŒH0Ds•1Ž-•i•‘’hd“”•• 
† ŒH0–K—D+,˜™BC 90∘1š›‡<ƒ/0œ7 
 
9."#$%90∘ 
"&'%v 𝐷 *••h•o•0 𝐴𝐵𝐶 žXL1𝐵𝐶 *Ÿ  
† "𝐵𝐴𝐶= 90∘ 
v G △𝐴𝐵𝐷 KL A xMNOH¡BCA △𝐴𝐶𝐷! D„… 
† "𝐷𝐴𝐷! = "𝐵𝐴𝐶= 90∘ 
 
10."#$%X 
!
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"&'%v ¢£<L-.DŽYLD¤¥Š‹¦}‘}•§¨ 
† 𝑚= 31𝑛= 2 
† (𝑚, 𝑛) ©} (3,2)1‰ª{X`« 
!
 
 
 
!
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1" #$%&
圆!"#$%&
D 
#'(&A")*+,-.,/01+)*-2345678 
B"/9:0;<=>?@A10B*-2345678 
C"CDEF>GH;IJ-K;I/01L;-2345678 
D";I+:0*MNO>G-2345PA- 
 
2" #$%&A 
#'(&∵ ,QR>*STH 90∘- 
∴ QU?TV+WX;T?TV- 
Y)IH 4- 
∴ ,QR>GO 4√2" 
24 A 
 
3" #$%&A 
#'(&∵;<𝑙Z)IH𝑟>⊙𝑂[\-]@𝑂^;<𝑙>_`H 5 - 
∴𝑟> 5- 
24aA" 
 
4" #$%&B 
 
 
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#'(&bc-de𝑂𝐴f𝑂𝐵-𝑂𝐺8 
∵ghV𝐴𝐵𝐶𝐷𝐸𝐹+hOH 2 >PghV- 
∴Δ𝑂𝐴𝐵+Wh?TV- 
∴𝑂𝐴= 𝐴𝐵= 2- 
∴𝑂𝐺= 𝑂𝐴⋅sin 60∘= 2 × √#
$ = √3- 
∴hOH 2 >PghV>ij*>)IH√3" 
24aB" 
 
5" #$%&D 
#'(&𝑆=
$%&'×)
#*&
= 6𝜋𝑐𝑚$- 
24 D" 
 
6" #$%&D 
#'(&kclm-*>noH 4𝜋 -PpV>RT<OqWr*>;I
H 4-QsRt>hOH 2√2-uPpV>noH 8-vwcV>Rxy-z
{noH 
4𝜋−8
4 |'U 𝜋−2" 
24 D 
 
7" #$%&5√3 
#'(& ∵𝐷 + 𝐴𝐵 M@ 
de 𝐶、𝐷  
∴𝐶𝐷=
/
$ 𝐴𝐵= 5  
∴𝐶𝐵= 𝐶𝐷= 5  
∵∠𝐶= 90∘  
∴𝐴𝐶= 5√3 
 
 
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8" #$%& 130°  
#'(&bc,∠𝐷H,𝐴𝐶QR>*}T, 
 
∵∠𝐷=
/
$ ∠𝐴𝑂𝐶, 
~∠𝐴𝑂𝐶= 100°, 
∴∠𝐷= 50° 
∵∠𝐷+ ∠𝐴𝐵𝐶= 180°,∠𝐴𝐵𝐶= 180° −50° = 130° 
 
9" #$%&√3 
#'(&de𝑂𝐵，∵𝑂𝐴f𝑂𝐵 •+)I- 
∴𝑂𝐵= 𝑂𝐴= 𝐴𝑃= 1- 
Y ∵𝑃𝐶 Z ⊙𝑂 [jr 𝐵 @- 
∴𝑂𝐵⊥𝑃𝐵- 
r+9 Rt △𝑃𝐵𝑂 M-𝑂𝐵= 1-𝑂𝑃= 2- 
∴𝑃𝐵= √2$−1$ = √3" 
2$%Ha√3" 
 
10"#$%&65∘ 
#'(&∵∠𝐴=
/
$ ∠1 =
/
$ × 130∘= 65∘- 
 ∴∠𝐶𝐷𝐸= ∠𝐴= 65∘" 
 
11"#$%&72 
#'(&∠𝐴𝑂𝐷=
$
0 × 360° = 144°- 
∴∠𝐴𝐵𝐷=
/
$ ∠𝐴𝑂𝐷=
/
$ × 144° = 72° 
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12"#$%&20𝜋 
#'(&∵*€>•+3𝑐𝑚-‚<O5𝑐𝑚- 
∴ƒ„1…U*€>†n)IH4𝑐𝑚- 
∴*€>‡no= 𝜋× 4 × 5 = 20𝜋𝑐𝑚$" 
 
13"#$%&ˆ 
#'(& 9 ⊙𝑂 M-𝐴𝐶
⏜
= 𝐵𝐷
⏜
- 
 ∵𝐵𝐶
⏜
= 𝐵𝐶
⏜
- 
 ∴𝐴𝐵
⏜
= 𝐶𝐷
⏜
- 
 ∴𝐴𝐵= 𝐶𝐷" 
 
14"#$%&8 ‰ 
#'(& 
 
ŠH‹q 𝐴𝐵= 24 m-ŒQ9*)IH 13 m 
•O 𝐶𝐷 ^ 𝑂-ŽU 𝑂𝐶= 𝑂𝐴-• 𝑂 H*S 
• 𝐴𝐷=
/
$ 𝐴𝐵= 12 (‰) 
• 𝑂𝐴= 13 ‰ 
9 Rt △𝐴𝑂𝐷 M-𝐷𝑂= √𝑂𝐴$ −𝐴𝐷$ = 5 
•~UŒ• 𝐶𝐷= 𝐶𝑂−𝐷𝑂= 13 −5 = 8 ‰ 
 
 
 
!
!
5 
新东方 老师好 
New Oriental Education & 
Technology Group!
15"(1) #$%&∠𝐶𝐴𝐷= 35∘ 
#'(&∵𝐴𝐵 +)* 𝑂 >;I- 
∴∠𝐴𝐶𝐵= 90∘- 
Y∵𝑂𝐷//𝐵𝐶- 
∴∠𝐴𝐸𝑂= 90∘-u 𝑂𝐸⊥𝐴𝐶- 
∠𝐶𝐴𝐵= 90∘−∠𝐵= 90∘−70∘= 20∘"∠𝐴𝑂𝐷= ∠𝐵= 70∘" 
∵𝑂𝐴= 𝑂𝐷- 
∴∠𝐷𝐴𝑂= ∠𝐴𝐷𝑂=
/
$ (180∘−∠𝐴𝑂𝐷) =
/
$ (180∘−70∘) = 55∘- 
∴∠𝐶𝐴𝐷= ∠𝐷𝐴𝑂−∠𝐶𝐴𝐵= 55∘−20∘= 35∘" 
(2) #$%&𝐴𝐵=
$0
# 
#'(&∵𝑂𝐸⊥𝐴𝐶- 
∴𝐴𝐸= 𝐸𝐶- 
9;T △𝐴𝐵𝐶 M-𝐷𝐸= √𝐴𝐷$ −𝐴𝐸$ = √5$ −4$ = 3" 
‘ 𝑂𝐴= 𝑂𝐷= 𝑟-• 𝑂𝐸= 𝑟−3" 
9;T △𝐴𝑂𝐸 M-𝐴𝑂$ = 𝐴𝐸$ + 𝑂𝐸$-𝑟=
$0
* " 
Y ∵𝑂𝐴= 𝑂𝐵- 
∴𝐴𝐵=
$0
# " 
 
小学
初中
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!
1 
新东方 老师好 
New Oriental Education & 
Technology Group!
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New Oriental Education & 
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新东方 老师好 
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