在线求助关于因式分解的题目！满意加分！

来源：百度知道 编辑：UC知道 时间：2024/04/28 23:50:02
1.(1) (2x+3)^2-2x-3 (2) (2x-1)^2-x^2

(3) 2y(2+y)=-(y+2) (4) (y-1)^2+2y(y-1)

(1)
提取(2x+3)
=(2x+3)(2x+3-1)
=(2x+3)(2x+2)
=2(2x+3)(x+1)

(2)
平方差
=(2x-1+x)(2x-1-x)
=(3x-1)(x-1)

(3)
怎么多出个等号就当减号了
提取(y+2)
=(y+2)(2y-1)

(4)
提取(y-1)
=(y-1)(y-1+2y)
=(y-1)(3y-1)

(3)2y(2+y)=-(y+2)
2y(2+y)+(y+2)=0
(2+y)(2y+1)=0
所以y=-2或-2分之1

(1) (2x+3)^2-2x-3
解：原式=(2x+3)^2-(2x+3)
=(2x+3)*(2x+3)-(2x+3)
=(2x+3)*(2x+3-1)
=(2x+3)*(2x+2)

(2) (2x-1)^2-x^2
解：原式=(2x-1)*(2x-1)-x^2
=4x^2-4x+1-x^2
=3x^2-4x+1
=（3x-1)(x-1)

(3) 怎么有=号，怎么因式分解？

(4) (y-1)^2+2y(y-1)
解：原式=(y-1)*(y-1)+2y(y-1)
=(3y-1)*(y-1)

要加分哦！~

1.(2x+3)(2x+2)
2.(x-1)(3x-1)
3.(2y-1)(y+2)
4.(3y-1)(y-1)