解方程x^2+3x=1
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第一题题目是不是不完整呀？
(2)x/(x-2)-(1-x^2)/(x^2-5x+6)=2x/(x-3)
x/(x-2)-(1-x^2)/[(x-2)(x-3)]=2x/(x-3)
[x(x-3)-2x(x-2)]/[(x-2)(x-3)]=(1-x^2)/[(x-2)(x-3)]
(x^2-3x-2x^2+4x)/[(x-2)(x-3)]=(1-x^2)/[(x-2)(x-3)]
(x-x^2)/[(x-2)(x-3)]=(1-x^2)/[(x-2)(x-3)]
两边同乘以(x-2)(x-3)
(x≠2且x≠3)，所以
x-x^2=1-x^2
解得x=1
(2)x/(x-2)-(1-x^2)/(x^2-5x+6)=2x/(x-3)
x/(x-2)-(1-x^2)/[(x-2)(x-3)]=2x/(x-3)
[x(x-3)-2x(x-2)]/[(x-2)(x-3)]=(1-x^2)/[(x-2)(x-3)]
(x^2-3x-2x^2+4x)/[(x-2)(x-3)]=(1-x^2)/[(x-2)(x-3)]
(x-x^2)/[(x-2)(x-3)]=(1-x^2)/[(x-2)(x-3)]
两边同乘以(x-2)(x-3)
(x≠2且x≠3)，所以
x-x^2=1-x^2
解得x=1
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