已知y=(x2+2x+1)\(x2-1)/(x+1)\(x2-x)-x+1.试说明不论x为何值,y的值不变
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![已知y=(x2+2x+1)\(x2-1)/(x+1)\(x2-x)-x+1.试说明不论x为何值,y的值不变](/uploads/image/z/14650347-3-7.jpg?t=%E5%B7%B2%E7%9F%A5y%3D%28x2%2B2x%2B1%29%5C%28x2-1%29%2F%28x%2B1%29%5C%28x2-x%29-x%2B1.%E8%AF%95%E8%AF%B4%E6%98%8E%E4%B8%8D%E8%AE%BAx%E4%B8%BA%E4%BD%95%E5%80%BC%2Cy%E7%9A%84%E5%80%BC%E4%B8%8D%E5%8F%98)
已知y=(x2+2x+1)\(x2-1)/(x+1)\(x2-x)-x+1.试说明不论x为何值,y的值不变
已知y=(x2+2x+1)\(x2-1)/(x+1)\(x2-x)-x+1.试说明不论x为何值,y的值不变
已知y=(x2+2x+1)\(x2-1)/(x+1)\(x2-x)-x+1.试说明不论x为何值,y的值不变
y=[(x^2+2x+1)/(x^2-1)]/[(x+1)/(x^2-x)]-x+1
=(x+1)^2/(x-1)(x+1)/[(x+1)/x(x-1)]-x+1
=(x+1)/(x-1)*x(x-1)/(x+1)-x+1
=x-x+1
=1
所以不论x为何值,y的值不变 ,不过x≠1且x≠-1且x≠0
y=(x+1)\(x-1)/(x+1)\(x-1)*x-(x-1)
=x-x+1
=1