作业帮初二两道分式计算.要计算过程.
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/04 07:29:10
初二两道分式计算.
要计算过程.
要计算过程.
计算什么呢?我就当因式分解了.
7.原式=[(x^2+5x+6)-1]/(x^2+5x+6) - [(x^2+8x+15)-1]/(x^2+8x+15)
=1 - 1/(x^2+5x+6) - 1 + 1/(x^2+8x+15)
=-1/(x+2)(x+3) + 1/(x+3)(x+5)
=-(x+5)/(x+2)(x+3)(x+5) + (x+2)/(x+2)(x+3)(x+5)
=[-(x+5)+(x+2)] / (x+2)(x+3)(x+5)
=-3/ (x+2)(x+3)(x+5)
8.原式={[x - 1/(x-y)] + [x + 1/(x-y)]}{[x - 1/(x-y)] - [x + 1/(x-y)]}
=(2x)[-2 /(x-y)]
=-4x/(x-y)
7.原式=[(x^2+5x+6)-1]/(x^2+5x+6) - [(x^2+8x+15)-1]/(x^2+8x+15)
=1 - 1/(x^2+5x+6) - 1 + 1/(x^2+8x+15)
=-1/(x+2)(x+3) + 1/(x+3)(x+5)
=-(x+5)/(x+2)(x+3)(x+5) + (x+2)/(x+2)(x+3)(x+5)
=[-(x+5)+(x+2)] / (x+2)(x+3)(x+5)
=-3/ (x+2)(x+3)(x+5)
8.原式={[x - 1/(x-y)] + [x + 1/(x-y)]}{[x - 1/(x-y)] - [x + 1/(x-y)]}
=(2x)[-2 /(x-y)]
=-4x/(x-y)