2/(x-1)(x-3)-1/(x-2)(x-3)+3/(x+2)(x-1) 用多种方法计算

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解决时间 2021-05-15 13:26

提问者网友：杀生予夺
2021-05-14 13:06

2/(x-1)(x-3)-1/(x-2)(x-3)+3/(x+2)(x-1) 用多种方法计算
要有过程.

最佳答案

五星知识达人网友：归鹤鸣
2021-05-14 13:22

（一）
2/[(x-1)(x-3)]-1/[(x-2)(x-3)]+3/[(x+2)(x-1) ]
=[2(x-2)-(x-1)]/[(x-1)(x-2)(x-3)]+3/[(x+2)(x-1)]
=(2x-4-x+1))/[(x-1)(x-2)(x-3)]+3/[(x+2)(x-1)]
=(x-3)/[(x-1)(x-2)(x-3)]+3/[(x+2)(x-1)]
=1/[(x-1)(x-2)]+3/[(x+2)(x-1)]
=[(x+2)+3(x-2)]/[(x+2)(x-1)(x-2)]
=(x+2+3x-6)/[(x+2)(x-1)(x-2)]
=(4x-4)/[(x+2)(x-1)(x-2)]
=4(x-1)/[(x+2)(x-1)(x-2)]
=4/[(x+2)(x-2)]
（二）
2/[(x-1)(x-3)]-1/[(x-2)(x-3)]+3/[(x+2)(x-1) ]
=2/[(x-1)(x-3)]+3/[(x+2)(x-1) ]-1/[(x-2)(x-3)]
=[2(x+2)+3(x-3)]/[(x+2)(x-1)(x-3)]-1/[(x-2)(x-3)]
=(2x+4+3x-9)/[(x+2)(x-1)(x-3)]-1/[(x-2)(x-3)]
=(5x-5)/[(x+2)(x-1)(x-3)]-1/[(x-2)(x-3)]
=5(x-1)/[(x+2)(x-1)(x-3)]-1/[(x-2)(x-3)]
=5/[(x+2)(x-3)]-1/[(x-2)(x-3)]
=[5(x-2)-(x+2)]/[(x+2)(x-2)(x-3)]
=(5x-10-x-2)/[(x+2)(x-2)(x-3)]
=(4x-12)/[(x+2)(x-2)(x-3)]
=4(x-3)/[(x+2)(x-2)(x-3)]
=4/[(x+2)(x-2)]

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