分式方程。x/(x-2)+(x-9)/(x-7)=(x+1)/(x-1)+(x-8)/(x-6)
2个回答
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(x-9)/(x-7)-(x+1)/(x-1)=(x-8)/(x-6)-x/(x-2)
[(x-9)(x-1)-(x+1)(x-7)]/[(x-7)(x-1)]=[(x-8)(x-2)-x(x-6)]/[(x-6)(x-2)]
[x^2-10x+9-x^2+6x+7]/(x^2-8x+7]=(x^2-10x+16-x^2+6x]/[x^2-8x+12]
(-4x+16)/(x^2-8x+7)=(-4x+16)/(x^2-8x+12)
所以-4x+16=0,
得:x=4
经检验为原方程的根
[(x-9)(x-1)-(x+1)(x-7)]/[(x-7)(x-1)]=[(x-8)(x-2)-x(x-6)]/[(x-6)(x-2)]
[x^2-10x+9-x^2+6x+7]/(x^2-8x+7]=(x^2-10x+16-x^2+6x]/[x^2-8x+12]
(-4x+16)/(x^2-8x+7)=(-4x+16)/(x^2-8x+12)
所以-4x+16=0,
得:x=4
经检验为原方程的根
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