已知x+1/x=2,求求分式(x²+2x+1)/(4x²-7x+4)的值
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已知x+1/x=2,求求分式(x²+2x+1)/(4x²-7x+4)的值
![已知x+1/x=2,求求分式(x²+2x+1)/(4x²-7x+4)的值](/uploads/image/z/6336221-5-1.jpg?t=%E5%B7%B2%E7%9F%A5x%2B1%2Fx%3D2%2C%E6%B1%82%E6%B1%82%E5%88%86%E5%BC%8F%EF%BC%88x%26%23178%3B%2B2x%2B1%EF%BC%89%2F%EF%BC%884x%26%23178%3B-7x%2B4%EF%BC%89%E7%9A%84%E5%80%BC)
x+1/x=2
两边同乘以x,并移项,得x²-2x+1=0
即(x-1)²=0
解得x=1
代入分式,
(x²+2x+1)/(4x²-7x+4)
=(1+2+1)/(4-7+4)
=4
两边同乘以x,并移项,得x²-2x+1=0
即(x-1)²=0
解得x=1
代入分式,
(x²+2x+1)/(4x²-7x+4)
=(1+2+1)/(4-7+4)
=4
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