整式乘除与因式分解的初一题

问题描述:

整式乘除与因式分解的初一题
1.化简
(1)(2a+1)(-a-2)+a(2-a)
(2)[(x+y)²-(x+y)(x-y)]÷(-2-y)
2.分解因式
-2x³+8xy²
3.已知x²-2x=2,求代数式(x-1)²+(x+3)(x-3)+(x-3)(x-1)的值
1个回答 分类:数学 2014-12-07

问题解答:

我来补答
1. 化简
(1)
(2a+1)(-a-2)+a(2-a)
= (-2a² -5a -2) +(2a -a²)
=-3a²-3a-2
(2)
[(x+y)²-(x+y)(x-y)]÷(-2-y)
= [(x² +2xy +y²) -(x² -y²)] ÷ [-(y+2)]
= -[2y² +2xy] ÷ (y+2)
= -(2y² +2xy)/(y+2)
=2xy/-y-2
2. 分解因式
-2x³+8xy²
= -2x(x² -4y²)
= -2x(x+2y)(x-2y)
3.已知x²-2x=2,求代数式(x-1)²+(x+3)(x-3)+(x-3)(x-1)的值
(x-1)²+(x+3)(x-3)+(x-3)(x-1)
= (x² -2x +1) +(x² -9) +(x² -4x +3)
= 3x² -6x -5
= 3(x² -2) -5
当x²-2x=2时
原式
= 3*2 -5
= 1.
 
 
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