What is the` (m-1)^(th)" root of "[(a^(m))^(m)-(1/m)]^(1/(m+1))` ?

Simply {(a^m)^(m-1/m)}^(1/(m+1))

Find square root of (m+(1)/(m))^(2)-4(m-(1)/(m)):

If m-(1)/(m)=5 then find m^(2)-(1)/(m^(2))

If the m^(th) term of an AP be (1)/(n) and n^(th) term (1)/(m) be , then show that its (mn)^(th) term is 1 .

Write the nth term of the A.P. (1)/(m) , (1+m)/(m) , (1+2m)/(m) ,…

The nth terms of an A.P. (1)/(m),(m+1)/(m),(2m+1)/(m),... is:

Show that the distance between (1,-1) and ((2m^2)/(1+m^2),(1-m)^2/(1+m^2)) is not depend on m

If m_(1)andm_(2) satisfy the relation "^(m+5)P_(m+1)=11/(2) (m-1) ("^(m+3)P_(m)) then m_(1)+m_(2) is equal to

If I(m,n)=int_0^1 t^m(1+t)^n.dt , then the expression for I(m,n) in terms of I(m+1,n-1) is: (a) (2^(n))/(m+1)-n/(m+1)I(m+1,n-1) (b) n/(m+1)I(m+1,n-1) (c) (2^(n))/(m+1)+n/(m+1)I(m+1,n-1) (d) m/(m+1)I(m+1,n-1)