Question

[2/m - 1/n] ÷ [m平方+n平方/n - 5n]×[m/2n + 2n/m + 2] 要过程

[2/m - 1/n] ÷ [m平方+n平方/n - 5n]×[m/2n + 2n/m + 2] 要过程

Answer 1

原式=(2n-m)/mn÷[(m²+n²-5n²)/n]×[(m²+4n²+4mn)/2mn]
=(2n-m)/mn×n/(m²-4n²)×(m+2n)²/2mn
=-(m-2n)/mn×n/(m+2n)(m-2n)×(m+2n)²/2mn
=-1/m(m+2n)×(m+2n)²/2mn
=-(m+2n)/(2m²n)

Answer 2

解：[2/m - 1/n] ÷ [m²+n²/n - 5n]×[m/2n + 2n/m + 2]
=（2n-m）/mn÷【（m²-4n²）/n】*【（m²+4n²+4mn）/(2mn)】
= -(m-2n）/mn*【（m+2n)²/(2mn)】*n/【(m+2n)(m-2n)】
= - (m+2n)/(2m²n)

Answer 3

（2/m-1/n）÷[(m²+n²)/n-5n]×(m/2n + 2n/m + 2) =-(m+2n)/(2m²n)
具体解题过程：
（2/m-1/n）÷[(m²+n²)/n-5n]×(m/2n + 2n/m + 2)
=(2n-m)/mn÷[(m²+n²-5n²)/n]×[(m²+4n²+4mn)/2mn]
=(2n-m)/mn×n/(m²-4n²)×(m+2n)²/2mn
=-(m-2n)/mn×n/(m+2n)(m-2n)×(m+2n)²/2mn
=-1/m(m+2n)×(m+2n)²/2mn
=-(m+2n)/(2m²n)