计算:(m+2n)/(n-m)+n/(m-n)-2m(n-m)和[(x+2)/(x×x-2x)-(x-1)/(x×x-4x+4)]÷(x-4)/x

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计算:(m+2n)/(n-m)+n/(m-n)-2m(n-m)和[(x+2)/(x×x-2x)-(x-1)/(x×x-4x+4)]÷(x-4)/x

计算:(m+2n)/(n-m)+n/(m-n)-2m(n-m)和[(x+2)/(x×x-2x)-(x-1)/(x×x-4x+4)]÷(x-4)/x
计算:(m+2n)/(n-m)+n/(m-n)-2m(n-m)和[(x+2)/(x×x-2x)-(x-1)/(x×x-4x+4)]÷(x-4)/x

计算:(m+2n)/(n-m)+n/(m-n)-2m(n-m)和[(x+2)/(x×x-2x)-(x-1)/(x×x-4x+4)]÷(x-4)/x
(m+2n)/(n-m)+n/(m-n)-2m(n-m)
=(m+2n-n-2m)/(n-m)
=(n-m)/(n-m)
=1
[(x+2)/(x×x-2x)-(x-1)/(x×x-4x+4)]÷(x-4)/x
={(x+2)/[x(x-2)]-(x-1)/(x-2)²}÷(x-4)/x
={[(x+2)(x-2)-x(x-1)]/[x(x-2)²]}÷(x-4)/x
=(x-4)/[x(x-2)²]* x/(x-4)
=1/(x-2)²

第一题:1
第二题:1/(x-2)(x-2)