已知x²+x-1=0,求x(1-2\(1-x))÷(x+1)-[x(x²-1)]\(x²-2x+1)的值.
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![已知x²+x-1=0,求x(1-2\(1-x))÷(x+1)-[x(x²-1)]\(x²-2x+1)的值.](/uploads/image/z/8625509-53-9.jpg?t=%E5%B7%B2%E7%9F%A5x%26%23178%3B%2Bx-1%3D0%2C%E6%B1%82x%281-2%5C%281-x%29%29%C3%B7%28x%2B1%29-%5Bx%28x%26%23178%3B-1%29%5D%5C%28x%26%23178%3B-2x%2B1%29%E7%9A%84%E5%80%BC.)
已知x²+x-1=0,求x(1-2\(1-x))÷(x+1)-[x(x²-1)]\(x²-2x+1)的值.
已知x²+x-1=0,求x(1-2\(1-x))÷(x+1)-[x(x²-1)]\(x²-2x+1)的值.
已知x²+x-1=0,求x(1-2\(1-x))÷(x+1)-[x(x²-1)]\(x²-2x+1)的值.
化简得:X/(X-1)-(X3-X)/(X-1)2
通分得:X2(I-X)/(X-I)2
再化简:X2/(1-X)
又X2=1-X
所以答案 1
我不会打X的平方,所以写的X2