y=(x-3)/(x-2)yx-2y=x-3(y-1)x=2y-3x=(2y-3)/(y-1)∴y≠1 问:为什么在x=(2y-3)/(y-1)这一步才y≠1 ,而不在(y-1)x=2y-3时讨论y是否等于1..依据是什么?
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![y=(x-3)/(x-2)yx-2y=x-3(y-1)x=2y-3x=(2y-3)/(y-1)∴y≠1 问:为什么在x=(2y-3)/(y-1)这一步才y≠1 ,而不在(y-1)x=2y-3时讨论y是否等于1..依据是什么?](/uploads/image/z/14550883-43-3.jpg?t=y%3D%28x-3%29%2F%28x-2%29yx-2y%3Dx-3%28y-1%29x%3D2y-3x%3D%282y-3%29%2F%28y-1%29%E2%88%B4y%E2%89%A01+%E9%97%AE%EF%BC%9A%E4%B8%BA%E4%BB%80%E4%B9%88%E5%9C%A8x%3D%282y-3%29%2F%28y-1%29%E8%BF%99%E4%B8%80%E6%AD%A5%E6%89%8Dy%E2%89%A01+%2C%E8%80%8C%E4%B8%8D%E5%9C%A8%28y-1%29x%3D2y-3%E6%97%B6%E8%AE%A8%E8%AE%BAy%E6%98%AF%E5%90%A6%E7%AD%89%E4%BA%8E1..%E4%BE%9D%E6%8D%AE%E6%98%AF%E4%BB%80%E4%B9%88%3F)
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