1、待定系数法.
2x²-3xy-2y²=(x-2y)(2x+y)
故设
2x²-3xy-2y²+3x+4y-2 =(x-2y+m)(2x+y-n)=2x²-3xy-2y²+(2m-n)x+(m+2n)y-mn
得:
2m-n=3
m+2n=4
-mn=-2
解得:m=2,n=1
所以2x²-3xy-2y²+3x+4y-2 =(x-2y+2)(2x+y-1)
2、思路同(1)
由10x²-27xy-28y²=(2x-7y)(5x+4y)
设10x²-27xy-28y²-x+25y-3=(2x-7y+m)(5x+4y+n)=10x²-27xy-28y²+(2n+5m)x+(-7n+4m)y-mn
可得:
5m+2n=-1
4m-7n=25
-mn=-3
解得:m=1,n=-3
从而
10x²-27xy-28y²-x+25y-3 =(2x-7y+1)(5x+4y-3)
3、题有问题吧?
因式化解 2x²-3xy-2y²+3x+4y-2 10x²-27xy-28y²-x+25-3 5x^5-x^4+x^3-x²+x-1
因式化解 2x²-3xy-2y²+3x+4y-2 10x²-27xy-28y²-x+25-3 5x^5-x^4+x^3-x²+x-1
数学人气:205 ℃时间:2019-10-14 06:30:57
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