内容正文:
)yg6a+7+2(02xy-6xy
练习19整式的乘除(1)
(8)-12x3y3+20z2y
1.(1)i6
89(42(6)号
(6)-1
2.-7ry2+21ry2)5a'6-81w
(3)0
2.(1)0.04x2-0.09(2)-2a20(3)m4-20m3-12m
1
④zy-8y(6)a-a6+a6-1a6+
(02y-9m2(6)-3x+2y-日
(6)a2-3ab-6b2(7)-3a2b-3ab2
a2b-4a2b2(6)-x8-10x+5
(8)c2+b2+d2+2bd-2bc-2cd-a2
3-9ry-3y(2a6t-12a6+号e6
(012-7。(10)-5y-2y+2
(3)3x2-4ax-3x(4)-x3+9x2-4
(5)x2m-4y2m+4y”-1
1.解:1)原式=20y-8x,当x=1y=号时,原式=2
练习16整式的乘法(2)
1.2-平ya12:
(2)原式=9-5,当x=一子时,原式=-8
练习20整式的乘除(2)
(4)-9a7b8(5)2x2ym
1.(1)-2m(2)-4x8y(3)9a4(4)-a
(6)29ab-10a2-10b2
2.(1)x6y6(2)6a8(3)-4mn3(4)-ab2+ab+2
62女-2y264
(5)-4a2+9a(6)6x3-x2+2x+2(7)-4a2b
2.(1)4x2-12x(2)-5ab3(3)-2xy
(8125r2-gy96n++1-5
(4)4m*n°三受m9n0(5)=40m+15m-)
(10)5.x2+6x-1
(6)2m+8(7)4b2(8)-2xy2+3y-1
3.5x3+5x2-20x-20
(9)-12x3y+24z2y2+3xy
练习17平方差公式与完全平方公式
(10)-8y2+4xy+2x-2
1.(1)4x2-1(2)25a2b2-1(3)m2n2+2mm+1
3.解:1)原式=4-4,当x=-2y=7时,原式
(4)
4-9m2(5)16x2-9y(6)4a2-9b2-6b-1
-10.(2)原式=5ab-62-3,当0=-号,6=2时,
2.(1)1(2)3999
8
(3)1(4)5
原式=一9,
4.解:原式=(x+1)(x-1)(x2+1)(x4+1)…(x208+
05a'-10ab+5b2(2)m2+4m(3)2xy-7
1)=(x2-1)(x2+1)(x4+1)…(x2048+1)=(x4-
(4)0(5)25-58m
1)(x4+1)…(x2048十1)=(x8-1)…(x208+1)=
42-台-y+1209w-+22-
x4096-1.
练习21整式的乘除(3)
(3)9y2-6y2+x2-4z2-4x-1
1.(1)-a(2)x18y8(3)-1(4)2x8y2
练习18整式的除法
1.1-4adc2Iwy2-1③号a1④-7min
(63a6-a6(6)-号9(-40
(8)-(x-y)2
(5)8.x(6)-b(7)6a2-3a-1(8)-16a8
2.(1)-3y3+2xy2+4(2)-8ab3-8a9(3)-22
2.(1)y-3x(2)-4x3y2(3)-1+2ab-3a2b
(4)-2a(5)9x2-y2+4y-4
(0x-2y(510w6)2a6
(6)-6a4+7a3+11a2-a-2
(7)-15.x2-y2+10xy(8)-8x2+99y2
《7)9a2+ga-3b(8)a+b2(9)a0
(9)-y2(10)-2x4+9x3-9x2+13x-3
032-号-1
(11)2x3”-2(12)24x-15
3.解:原式=4a2-2ab,当a=2,b=1时,原式=12.
3.(1)10(2)-15(3)4或
1
·10·六年级数学下LJ
同行学案学练测计算高手
练习21整式的乘除(3)
(难度等级:★★★★)
V基础通关
(4)4a2b(-2ab)2÷(-0.5a3b)
1.计算.
(1)(-a)1021÷(-a)1020=
(2)(x2y2)2(x3y3)3=
(3)(-a3)4÷(-a4)3=
(4)(x2y3)4+(-x)8·(y)2=
(6)(号ab2-2ab)·2ab=
(5)(3x+y-2)(3x-y+2)
(6(2g)÷(-8pg)=
((3a+48)(482-3a)=
(8)(x一y)7÷(y-x)3÷(y-x)2=
2.计算.
(6)(3a2+a-1)·(-2a2+3a+2)
(1)(15x3y5-10x4y4-20x3y2)÷(-5x3y2)
(7)(3x-2y)(y-3x)-(2x-y)(3x+y)
(2)(-2a2b)3+8(a2)2·(-a)2·(-a)3
(8)(2x十+3y)2-(4x-9y)(4x十9y)+(2x-3y)2
(3(-4红2)·(日xy)÷(2xy2)月
·23·
六年级数学下L
同行学案学练测计算高手
(9)(6x3y-2x2y2-2xy3)÷(-2xy)-
V能力通关
(3x+2y)(y-x)
3.已知m-n=4,mn=-3.
(1)计算:m2+n2.
(10)(2x2-x+3)(-x2+4x-1)
(2)求(m2-4)(n2-4)的值.
(11)(x3m+1)(x3m-1)-(x3m-1)2
(3)求8m·32÷4m+m的值.
(12)(2x+1)2(-2x"+1)2-16(x”+
1)2(x"-1)2
·24·