laws of rational exponent (v)` (ab)^p` =` a^p b^p` (vi)`(a/b)^p` =` a^p/b^p` (vii) `a^(p/q) = (a^p)^(1/q) = (a^q)^(1/p)`

laws of Rational exponents are same as real exponents.(i) a^(p)xx a^(q)=a^(p+q)( ii) (a^(p))/(a^(q))=a^(p-q) (iii) (a^(p))^(q)=a^(pq)( iv )a^(-q)=(1)/(a^(q))

Prove that a^(p)b^(q)<((ap+bq)/(p+q))^(p+q)

If in a g.P. { t_(n)) it is given that t_(p+q) =a and t_(p-q) = b then : t_(p) = (A) (ab)^(1/2) (B) (ab)^(1/3) (C) (ab)^(1/4) (D) none of these

Let abs(a)=abs(b)=2" and "p=a+b, q=a-b . If abs(p times q)=2(k-(a.b)^(2))^(1//2) then

If p and q and q are two statements , prove that (i) ~( p vee q) -= (~p) ^^ ( q) " " (ii) ~( p ^^ q) = (~p) v(~q)

(2) If a=x^(q+r)*y^(p) , b=x^(r+p)*y^(q) , c=x^(p+q)y^(r) , show that a^(q-r)b^(r-p)c^(p-q)=1

The roots of the equation (q-r)x^2+(r-p)x+p-q=0 are (A) (r-p)/(q-r),1 (B) (p-q)/(q-r),1 (C) (q-r)/(p-q),1 (D) (r-p)/(p-q),1