先化简,再求值:(x-1/x)-(x-2/x+1)除以2x^2-x/x^2+2x+1,其中x满足x^2-x-1=0.
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![先化简,再求值:(x-1/x)-(x-2/x+1)除以2x^2-x/x^2+2x+1,其中x满足x^2-x-1=0.](/uploads/image/z/757142-62-2.jpg?t=%E5%85%88%E5%8C%96%E7%AE%80%2C%E5%86%8D%E6%B1%82%E5%80%BC%EF%BC%9A%EF%BC%88x-1%2Fx%29-%28x-2%2Fx%2B1%29%E9%99%A4%E4%BB%A52x%5E2-x%2Fx%5E2%2B2x%2B1%2C%E5%85%B6%E4%B8%ADx%E6%BB%A1%E8%B6%B3x%5E2-x-1%3D0.)
先化简,再求值:(x-1/x)-(x-2/x+1)除以2x^2-x/x^2+2x+1,其中x满足x^2-x-1=0.
先化简,再求值:(x-1/x)-(x-2/x+1)除以2x^2-x/x^2+2x+1,其中x满足x^2-x-1=0.
先化简,再求值:(x-1/x)-(x-2/x+1)除以2x^2-x/x^2+2x+1,其中x满足x^2-x-1=0.
[(x-1)/x-(x-2)/(x+1)]÷[(2x^2-x)/(x^2+2x+1)]
=[(x-1)(x+1)/x(x+1)-x(x-2)/x(x+1)]÷[x(2x-1)/(x+1)²]
={[(x-1)(x+1)-x(x-2)]/[x(x+1)]}×[(x+1)²/x(2x-1)]
=[(2x-1)(x+1)]/[x²(2x-1)]
=(x+1)/x²
=1
上式最后一步利用条件:x^2-x-1=0
可化为:x^2=x+1