1/x+2-1/x+3-1/x+4+1/x+5=0

来源：百度知道 编辑：UC知道 时间：2024/06/17 21:05:04
RT 1/x+2-1/x+3-1/x+4+1/x+5=0
有过程的 谢谢

移项得:1/(x+2)+1/(x+5)=1/(x+3)+1/(x+4);两边各通分得:[(x+2)+(x+5)]/(x+2)(x+5)=[(x+3)+(x+4)]/(x+3)(x+4);即:(2x+7)/(x^2+7x+10)=(2x+7)/(x^2+7x+12),分子相同,分母不相等,所以分子同为零2x+7=0;解得:x=-7/2

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1/x+2-1/x+3-1/x+4+1/x+5=0
1/x+2-1/x+3=1/x+4-1/x+5
[(x+3)-(x+2)]/[(x+2)(x+3)]=[(x+5)-(x+4)]/[(x+4)(x+5)]
1/[(x+2)(x+3)]=1/[(x+4)(x+5)]
(x+2)(x+3)=(x+4)(x+5)
x^2+5x+6=x^2+9x+20
4x=-14
x=-3.5

移项得:1/(x+2)+1/(x+5)=1/(x+3)+1/(x+4);两边各通分得:[(x+2)+(x+5)]/(x+2)(x+5)=[(x+3)+(x+4)]/(x+3)(x+4);即:(2x+7)/(x^2+7x+10)=(2x+7)/(x^2+7x+12),分子相同,分母不相等,所以分子同为零2x+7=0;解得:x=-7/2