题目
x+2/x+1-x+3/x+2-x-4/x-3+x-5/x-4如何分解因式
提问时间:2020-10-19
答案
原式=(x+2)/(x+1)-(x+3)/(x+2)-(x-4)/(x-3)+(x-5)/(x-4)
=[(x+2)*(x+2)-(x+3)*(x+1)]/(x+1)*(x+2)--[(x-4)*(x-4)-(x-3)*(x-5)]/(x-3)*(x-4)
=1/(x+1)*(x+2)-1/(x-3)*(x-4)
=[(x-3)*(x-4)-(x+1)*(x+2)]/[(x+1)*(x+2)*(x-3)*(x-4)]
=10(1-x)/[(x+1)*(x+2)*(x-3)*(x-4)]
=[(x+2)*(x+2)-(x+3)*(x+1)]/(x+1)*(x+2)--[(x-4)*(x-4)-(x-3)*(x-5)]/(x-3)*(x-4)
=1/(x+1)*(x+2)-1/(x-3)*(x-4)
=[(x-3)*(x-4)-(x+1)*(x+2)]/[(x+1)*(x+2)*(x-3)*(x-4)]
=10(1-x)/[(x+1)*(x+2)*(x-3)*(x-4)]
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