(3-2sqrt(2))/(3+2sqrt(2))

Rationalise the denominator of each of the following (3-2sqrt2)/(3+2sqrt2)

A=(3sqrt(2)-2)/(3sqrt(2)+2),B=(3sqrt(2)+2)/(3sqrt(2)-2), then A+B

After rationalising the denominator of (3 sqrt(2))/(3 sqrt(2) - 2 sqrt(2)) , we get the denominator as

If 1/(4-sqrt8)+(3+2 sqrt2)/(3-2 sqrt2)- (3-2 sqrt2)/(3+2 sqrt2)=a+b sqrt2 , then what is the value of (3a+4b) ? यदि 1/(4-sqrt8)+(3+2 sqrt2)/(3-2 sqrt2)- (3-2 sqrt2)/(3+2 sqrt2)=a+b sqrt2 तो (3a+4b) का मान ज्ञात कीजिए?

[(2-sqrt(3))/(2sqrt(2))]/[(2+sqrt(3))/(2sqrt(2))]

Simplify: (3sqrt(2)-2sqrt(2))/(3sqrt(2)+2sqrt(3))+(sqrt(12))/(sqrt(3)-sqrt(2)) (ii) (sqrt(5)+sqrt(3))/(sqrt(5)-sqrt(3))+(sqrt(5)-sqrt(3))/(sqrt(5)+sqrt(3))

Simplify: (i) (3sqrt(2)-2sqrt(2))/(3sqrt(2)+\ 2sqrt(3))+(sqrt(12))/(sqrt(3)-\ sqrt(2)) (ii) (sqrt(5)+\ sqrt(3))/(sqrt(5)-\ sqrt(3))+(sqrt(5)-\ sqrt(3))/(sqrt(5)+\ sqrt(3))

(3+sqrt(2))(2-sqrt(3))(3-sqrt(2))(2+sqrt(3))

The sum of two numbers is 6 times their G.M. Show that the numbers are in the ratio (3+2sqrt2):(3-2sqrt2)

The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3+2sqrt2):(3-2sqrt2)