If `a^2x^4+b^2y^4=c^6`, then the maximum value of `xy` is (a) `c^3/(2ab)` (b) `c^3/sqrt(2ab)` (c) `c^3/(ab)` (d) `c^3/sqrt(ab)`

If a^(2)x^(4)+b^(2)y^(4)=c^(6), then the maximum value of xy is (a) (c^(3))/(2ab) (b) (c^(3))/(sqrt(2ab)) (c) (c^(3))/(ab) (d) (c^(3))/(sqrt(ab))

If a^2x^4+b^2y^4=c^6, then the maximum value of x y is (a) (c^2)/(sqrt(a b)) (b) (c^3)/(a b) (c) (c^3)/(sqrt(2a b)) (d) (c^3)/(2a b)

If a^2x^4+b^2y^4=c^6, then the maximum value of x y is (c^2)/(sqrt(a b)) (b) (c^3)/(a b) (c^3)/(sqrt(2a b)) (d) (c^3)/(2a b)

If a^(2)x^(4)+b^(2)y^(4)=c^(6), then the maximum value of xy is (a) (c^(2))/(sqrt(ab)) (b) (c^(3))/(ab)(c^(3))/(sqrt(2ab)) (d) (c^(3))/(2ab)

Prove that the greatest value of xy is c^(3)/sqrt(2ab)* if a^(2)x^(4)+b^(2)y^(4)=c^(6).

If a/2=b/3 =c/4 =(3a-ab+4c)/p , then find the value of p.

If a=-2, b=-3, c = 2 then find the value of (ab)/(c) + 9 ?

If x:y=3:4land a:b=1:2, then the value of (2xa+yb)/(3yb-4xa) is (5)/(6) b.(6)/(7) c.(6)/(5) d.(7)/(6)

Factorise : 2ab^(2) c - 2a + 3b^(3) c - 3b - 4b^(2) c^(2) + 4c