" If "[mn][[m],[n]]=25" and "m

If [m,n][[mn]]=[25] and (m,n)

If the average of m numbers is n^(2) and that of n numbers is m^(2) ,then the average of (m+n) numbers is m-n (b) mn(c)m+n (d) (m)/(n)

Add: 3m^(2)n + 5mn^(2) - 7mn and 2m^(2)n - mn^(2) + mn

if m+n=5 and mn= 6, but (m^2 + n^2) (m^3 + n^3) = how much?

If m and n are two distinct numbers such that m >n, then prove that m^2-n^2 ,2mn and m^2+n^2 is a Pythagorean triplet.

If mn = 3(m+1) + n and m and n are integers, m could be any of the following values EXCEPT: