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份学霸智慧课堂数学八年级上册
(2)解:△CDE是等腰直角三角形.理由如下:
(2)x十y=3,.xy=2.
:R△ADE≌R△BEC,∴∠ADE=∠BEC
.x2+3xy十y
又∠ADE+∠AED=90°,,∴.∠BE+∠AED=90°
=(x十y)2+xy
,.∠DEC=180°-(∠BEC+∠AED)=90.
=32+2
又DE=EC,
=11.
.△CDE是等腰直角三角形,
22.解:2-2
23.(1)270°(2)220
=2(22-1)
(3)∠1+∠2=180+∠A
=2(24+1)(28-1)
(4)解:"△EFP是由△EFA折叠得到的,
=2(26+1)(28+1)(2-1)
∴.∠AFE=∠PFE,∠AEF=∠PEF,
=2(2+1)(28+1)(2+1)(2-1)
.∠1=180°-2∠AFE,∠2=180°-2∠AEF
=2(26+1)(28+1)×17×15.
∴.∠1+∠2=360°-2(∠AFE+∠AEF).
.这两个数是17.15.
又,'∠AFE+∠AEF=180°一∠A,
23.解:(1)2m2+5+2n2=(m+2n)(2m+n)
∴.∠1+∠2=360°-2×(180°-∠A)=2∠A.
(2)阴影部分的面积为a+:-之女-之以a+b)=之(a
第十四章测试卷
+6-ab)=2ta+b2-3ah1=合(8-3X12)=t.
1.C2.C3.B4.D5.A6.B7.D8.A9.B10.B
11.-212.a(6-2)213.814.5215.3
答:阴影部分的面积为14,
16.解:(1)原式=8.x5-x
(3)由题意,得m=12.5,2mm2十22=48,
=7x.
∴m+n2=24.
(2)原式=(a-9a)·a
,∴.(m十n)2=m十2十2mn=24十25=49.
=(-8a)·a
:m>n>0,.m十n=7.
=-8a3.
,.图中所有裁剪线(虚线部分)长之和为6(m+)=
(3)方法1:原式=(一xy)2十2·(-xy)·1十1日
42(cm).
=x'y-2xy+I.
答:图中所有裁剪线(虚线部分)长之和为42cm
方法2:原式=(1一xy)
第十五章测试卷
=1-2·xy·1+(xy)月
=1-2ry+ry,
1.A2.C3.B4.C5.A6.C7.A8.A9.A10.B
(4)原式=(9x2y-27.xy2)÷9xy
1.112.若13.-2143.4×10015.0或-4或-8
=9x2y2÷9x2y2-27xy2÷9.x2y
=y-3.x.
16.解:1)最简公分母为6y·27y6·3可6r了
1=3y12.x
17.解:(1)原式=a2(a-b)-(a一b)
x(+y)
=(a-b)(a-F)
=(a一b)2(a+b).
(2)原式x千一y一寸
(2)原式=(m一n)2-1
(3)d+a-1=4-a+1
1-a2-a3a3+a2-1
=(m一+1)(m一一1).
9ub
18.解:原式=2(-8xy十16y2)=2(x-4y).
忆.解:原式=·a十a-。方
3ab
当x=3,y=1时,
原式=2×(3-4×1)=2.
(2)原式=4=2+3.a+2(a-2
a-2
a+1
19.(1)(2x-3y11-(x-2y)(x+2y)
=a十2.
=4x{6xy+3yx2网第一步
18.解:原式=2a
2(a-2)
.(a-1)月
a+1(a-1)(a+1a-2
=3x2-6xy+第三步
(2)解:(2x-3y)-(x-2y)(x+2y)
=24_2(a-1
a十1a十1
=4x2-12xy+9y2-x2+4y
2
=3.x2-12xy十13y.
a+1
20.解:(1)(x2一m.x一n)(x一2)=x2-mx2-x-2x2+2nx
+2n=x3一(m+2)2+(21-n).x+2n.
当a=一2时,原式=-2干有一2
由题可知m十2=0,2m一n=0,
19.解:(1)方程两边乘x(x-2),得x2-3x+2+2x=x-2x,
解得m=一2,n=一4.
解得x=一2.
(2)原式=(m2+2mn+r2一2m-m2一8m)÷2m
检验:当x=一2时,x(x一2)≠0
=(7m2-8m)÷2m
.原分式