作业帮已知y=[(x^2+6x+9)/(x^2-9)]/[(x+3)/(x^2-3x)]-kx+3,不论x为任何有意义的值,y的值均不变求k的值
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![已知y=[(x^2+6x+9)/(x^2-9)]/[(x+3)/(x^2-3x)]-kx+3,不论x为任何有意义的值,y的值均不变求k的值](/uploads/image/z/4346044-52-4.jpg?t=%E5%B7%B2%E7%9F%A5y%3D%5B%28x%5E2%2B6x%2B9%29%2F%28x%5E2-9%29%5D%2F%5B%28x%2B3%29%2F%28x%5E2-3x%29%5D-kx%2B3%2C%E4%B8%8D%E8%AE%BAx%E4%B8%BA%E4%BB%BB%E4%BD%95%E6%9C%89%E6%84%8F%E4%B9%89%E7%9A%84%E5%80%BC%2Cy%E7%9A%84%E5%80%BC%E5%9D%87%E4%B8%8D%E5%8F%98%E6%B1%82k%E7%9A%84%E5%80%BC)
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已知y=[(x^2+6x+9)/(x^2-9)]/[(x+3)/(x^2-3x)]-kx+3,不论x为任何有意义的值,y的值均不变求k的值
已知y=[(x^2+6x+9)/(x^2-9)]/[(x+3)/(x^2-3x)]-kx+3,不论x为任何有意义的值,y的值均不变
求k的值
已知y=[(x^2+6x+9)/(x^2-9)]/[(x+3)/(x^2-3x)]-kx+3,不论x为任何有意义的值,y的值均不变求k的值
化简后得y=(1-k)x+3所以k=1