Question

计算:[-m^3n/(m-n)^2]^4*[(n^2-mn)/m]^3/m^4n^10*[(mn-n^2)/m]^6

Answer 1

[-m^3n/(m-n)^2]^4*[(n^2-mn)/m]^3/m^4n^10*[(mn-n^2)/m]^6

=m^12n^4/(m-n)^8*[n(n-m)/m]^3/m^4n^10*[n(m-n)/m]^6

=m^8/(m-n)^8*n^3(n-m)^3/m^3/n^6*n^6(m-n)^6/m^6

=-m^2/(m-n)^2*n^3(m-n)^3/m^3

=-n^3(m-n)/m