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

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

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)

If m^2+m^('2)+2mn'cos theta =1 n^2+n^('2)+2"nn"' cos theta=1 and mn +m'n'+(mn'+m'n)cos theta=0 then show that m^2+n^2=m^('2) +n^('2) =cosec^2 theta .

Prove that (mn)! Is divisible by (n!)^(m) " and" (m!)^(n) .