因式分解(x²+4)²-(2x+3)² 99²+99分之99²+199
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/11/11 13:11:57
因式分解(x²+4)²-(2x+3)² 99²+99分之99²+199 4a²b²-(2x+3)²
(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)
再问: 4a²b²-(2x+3)² 换成4a²b²-(a²+b²)²应该怎么算?
再答: =-[(a²+b²)²-(2ab)²] =-(a²+b²+2ab)(a²+b²-2ab) =-(a+b)²(a-b)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)
再问: 4a²b²-(2x+3)² 换成4a²b²-(a²+b²)²应该怎么算?
再答: =-[(a²+b²)²-(2ab)²] =-(a²+b²+2ab)(a²+b²-2ab) =-(a+b)²(a-b)²