问题标题:
因式分解(x²+4)²-(2x+3)²99²+99分之99²+1994a²b²-(2x+3)²
问题描述:
因式分解(x²+4)²-(2x+3)²99²+99分之99²+1994a²b²-(2x+3)²
丁宝回答:
(x²+4)²-(2x+3)²
=(x²+4+2x+3)(x²+4-2x-3)
=(x²+2x+7)(x²-2x+1)
=(x²+2x+7)(x-1)²
(99²+199)/(99²+99)
=(99²+2×99+1)/(99²+99)
=(99+1)²/[99×)99+1)]
=(99+1)*99
=99分之100
4a²b²-(2x+3)²
=(2ab+2x+3)(2ab-2x-3)
黄善钧回答:
4a²b²-(2x+3)²换成4a²b²-(a²+b²)²应该怎么算?
丁宝回答:
=-[(a²+b²)²-(2ab)²]=-(a²+b²+2ab)(a²+b²-2ab)=-(a+b)²(a-b)²
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