内容正文:
1021
强化提升专项②整式的综合运算
(3)5(6+1)(6+1)(6+1)(6+1)(6+1)+1.
类型1
整式的运算
1.计算:
D(侵))厂+(x-3)--31+(-12:
3.计算:
(1)a(a+2h)-(a+b):
(2)(-2xy)2-2xy·(ry):
(2(2aw-20+aw)=(-2a:
(3)(-3a)3+(-2a)÷(-a)5:
培优训练
(3)4aa-3h)-(3b-2a)(2a+3b):
(4)(-a2)2·a5+a"÷a-(-2a2)3.
(4)(3ab+4)2-(-4+3ab)2,
2.简便计算:
(1)0.25230×4m1×(-8)180X0.50:
(5)(x-2)(x2+2x+4):
(2)2020-4042×2020+20212:
0221
(6[x+2y)-(8x+3x-y)-5y]÷7
(4)(2021·深圳福田区月考)先化简,再求值:
[(a+2b)(a-2b)-(a-2b)]÷2b,其中a+21+
(6-1)2=0.
(7)(3+2a+b)(3-2a+b).
角度2整体代入化简求值
5.(1)已知x2-5x=14,求(x-1)(2x-1)-(x+1)
十1的值
类型2化简求值
角度1直接代入化简求值
4.(1)先化简,再求值:(x-2y)(x+4y)-(2r-y)(x
十y),其中x=-2,y=3.
培优训练
(2)(2021·杭州拱墅区期中)已知x一2x一2=0,
将下式先化简,再求值:(x一1)P一(x十3)(x-3)
-(x3)(x-1).
(2)(2021·福州仓山区期末)先化简,再求值:
(x-1)(.x+6)-(6.x+10.x3-12.x)÷2.x2,其
中x=-2
(3)(2021·杭州西湖区期中)先化简,再求值:
(m-4n)2-4n(3n-2m)-3(-2n+3m)(3m十
2n),其中13m2-8m2-6=0.
(3)(2021·茂名茂南区月考)先化简,再求值:
[(x+y)(x-y)-(x-y)2-y(x-2y)]÷2x
其中x=-2021,y=-2.即3+2ab=4,解得ab=
强化提升专项2整式的综合运算
d+8-3.ab-
1.解:(1)原式=2+1-3十1=1.
(2)原式=4xy-2x'y=2xy
∴a+6=(a2+)-2a6=3-2×(分)=82
(3)原式=-27a3+4a÷(-d°)=-27a+(-4a)=-31a
(4)原式=a'·a°+a°+8a'=a’+a’+8a'=10a.
3.解:因为(x一y)=2,所以一2xy+y=2.
因为a2十y=1,所以2xy=-1,
2解:0)原式=(什×4)×4×(-2)“×(侵)=1×4×
所以(x+y)=x+2y+y=1-1=0,
所以十y=0.
(-2x2)=1×4×(-1)=4X1=4
4.解:(a-b=10.∴a2-2ab+6=10.①
(2)原式=20203-2×2021×2020+2021=(2020-2021)°=
,(a+b)2=6,∴.a2+2ab+6=6.②@
(-1)2=1.
①一②,得一4ab=4,所以ab=一1.
(3》原式=(6-1)(6+1)(6+1)(6+1)(6+1)(61+1)+1=(6
①十②,得2a十2=16,所以a十6=8.
-1)(6+1)(6+1)(6+1)(6+1)+1=(6-1)(6+1)(6+1)·
5.解:.a--b=-10.(a-b)·c=-12,
(6+1)+1=(6-1)6+1D(66+1)+1=(6"-1D(6"+1)+1
.(a-b)°+c=[(a-b)-c]+2(a-b)·c=(-10)2+2×(-12)
6-1+1=6.
=76.
3.解:(1)原式=a2+2ab-a2-2ah-b=-6.
6解:“+上=4(+)广=16
(2)原式=(年aW-之aW+6)÷a6=3ab-2ab+1.
d+7-(e+)广-2=16-2=14
(3)原式=4a2-12ah-9+4a3=8a2-12ah-9h.
(x-士)=+-2=14-2=12
(4)原式=9a'b+24ab十16-(9ab-24ab+16)=9ab+24ab+16
-9ab+24ah-16=48ad.
7.解:(1)(a+b)2-(b-u)2=4ab
(5)原式=z+2.x2十4x-2x2-4x-8=x-8.
(2)图为(2r+y)2-(2r-y)-8ry
所以8ry=169一9,所以xy=20.
(6)原式=(r+4ry+4y-9r+y-5y)÷号x=(-8r+4
7整式的除法
÷宁r=-16r+8x
第1课时单项式除以单项式
(7)原式-[(3+)+2a[(3+)-2a]=(3+)2-4a=9+6M+-4a2
1.C2.D3.A4.A
4解:(1)原式=2+4y-2y-8