Rationalise the denominator of the following : `(a)(2)/(3sqrt(3))" "(b)(sqrt(40))/(sqrt(3))(c )" "(3+sqrt(2))/(4sqrt(2))" "(d)(16)/(sqrt(41)-5)" "(e)(2+sqrt(3))/(2-sqrt(3))`

Answer

N/a

Explanation

(a) `(2)/(3sqrt(3))=(2)/(3sqrt(3))xx(sqrt(3))/(sqrt(3))=(2sqrt(3))/(9)`
(b) `sqrt(10)/(sqrt(3))=(sqrt(4xx10))/(sqrt(3))xxsqrt(3)/(sqrt(3))=(2sqrt(30))/(3)`
( c) `(3+sqrt(2))/(4sqrt(2))=(3+sqrt(2))/(4sqrt(2))xx(sqrt(2))/(sqrt(2))=(3sqrt(2)+2)/(4(2))=(3sqrt(2)+2)/(8)`
`(16)/(sqrt(41)-5)=(16)/(sqrt(41)-5)xx(sqrt(41+5))/(sqrt(41+5))=(16sqrt(41)+5)/((sqrt(41))^(2)-(5)^(2))=(16(sqrt41+5))/(41-25)=(16(sqrt(41)+5))/(16)=sqrt(41)+5`
(e) `(2+sqrt(3))/(2-sqrt(3))=(2+sqrt(3))/(2-sqrt(3))xx(2+sqrt(3))/(2+sqrt(3))=((2)^(2)+(sqrt(3))^(2)+2(2)(sqrt(3)))/((2)^(2)-(sqrt(3))^(2))=(4+3+4sqrt(3))/(4-3)=(7+4sqrt(3))`
(f) `(sqrt(6))/(sqrt(2)+sqrt(3))=(sqrt(6))/(sqrt(2)+sqrt(3))xx(sqrt(3)-sqrt(2))/(sqrt(3)-sqrt(2))=(sqrt(18)-sqrt(12))/((sqrt(3))^(2)-(sqrt(2))^(2))=(3sqrt(2)-2sqrt(3))/(3-2)=3sqrt(2)-2sqrt(3)`
(g) `(sqrt(3)+sqrt(2))/(sqrt(3)-sqrt(2))=(sqrt(3)+sqrt(2))/(sqrt(3)-sqrt(2))xx(sqrt(3)+sqrt(2))/(sqrt(3)+sqrt(2))=((sqrt(3))^(2)+(sqrt2)^(2)+2(sqrt(3))(sqrt(2)))/((sqrt(3))^(2)-(sqrt(2))^(2))=(3+2+2sqrt(6))/(3-2)=2sqrt(6)`
(h) `(3sqrt(5)+sqrt(3))/(sqrt(5)-sqrt(3))=(3sqrt(5)+sqrt(3))/(sqrt(5)-sqrt(3))xx(sqrt(5)+sqrt(3))/(sqrt(5)+sqrt(3))=(3sqrt(5)(sqrt(5)+sqrt(3))+sqrt(3)(sqrt(5)+sqrt(3)))/((sqrt(5))^(2)-(sqrt3)^(2))=(15+3sqrt(15)+sqrt(15)+3)/(5-3)=(18)/(2)+(4sqrt(15))/(2)=9+2sqrt(15)`
(i) `(4sqrt(3)+5sqrt(2))/(sqrt(48)+sqrt(18))=(4sqrt(3)+5sqrt(2))/(sqrt(4xx4xx3)+sqrt(3xx3xx2))=(4sqrt(3)+5sqrt(2))/(4sqrt(3)+3sqrt(2))=(4sqrt(3)+5sqrt(2))/(4sqrt(3)+3sqrt(2))xx(4sqrt(3)-3sqrt(2))/(4sqrt(3)-3sqrt(2))`
`=(4sqrt(3)(4sqrt(3)-3sqrt(2))+5sqrt(2)(4sqrt(3)-3sqrt(2)))/((4sqrt(3))^(2)-(3sqrt(2))^(2))=(48-12sqrt(6)+20sqrt(6)-30)/(48-18)`
`=(18+8sqrt(6))/(30)=(2(9+4sqrt(6)))/(30)=(9+4sqrt(6))/15`