Find the values of a and b in each of the following :
`(a)(5+2sqrt3)/(7+4sqrt(3))=a-6sqrt(3)" "(b)(3-sqrt(5))/(3+2sqrt(5))=asqrt(5)-(19)/(11)`
`(c )(sqrt(2)+sqrt(3))/(3sqrt2-2sqrt(3))=2-bsqrt(6)" "(d)(7+sqrt(5))/(7-sqrt(5))-(7-sqrt(5))/(7+sqrt(5))=a+(7)/(11)sqrt(5b)`

Answer

N/a

Explanation

(a) `a-6sqrt(3)=(5+2sqrt(3))/(7+4sqrt(3))=(5+2sqrt(3))/(7+4sqrt(3))xx(7-4sqrt(3))/(7-4sqrt(3))`
`=(35-20sqrt(3)+14sqrt(3)-24)/((7)^(2)-(4sqrt(3))^(2))=(11-6sqrt(3))/(49-48)=11-6sqrt(3)`
Comparing, we get
a = 11
(b) `asqrt(5)-(19)/(11)=(3-sqrt(5))/(3+2sqrt(5))=(3-sqrt(5))/(3+2sqrt(5))xx(3-2sqrt(5))/(3-2sqrt(5))=(9-6sqrt(5)-3sqrt(5)+10)/((3)^(2)-(2sqrt(5))^(2))`
`=(-9sqrt(5)+19)/(-11)`
`=(-9sqrt(5))/(-11)+(19)/(-11)=(9)/(11)sqrt(5)-(19)/(11)`
Comparing, we get
`a=(9)/(11)`
(c ) `2-bsqrt(6)=(sqrt(2)+sqrt(3))/(3sqrt(2)-2sqrt(3))=(sqrt(2)+sqrt(3))/(3sqrt(2)-2sqrt(3))xx(3sqrt(2)+2sqrt(3))/(3sqrt(2)+2sqrt(3))`
`=(6+2sqrt(6)+3sqrt(6)+6)/((3sqrt(2))^(2)-(2sqrt(3))^(2))=(12+5sqrt(6))/(18-12)=(12)/(6)+(5sqrt(6))/(6)=2+(5)/(6)sqrt(6)`
Comparing, we get `-b=(5)/(6)" "rArr" "b=(-5)/(6)`
(d) `a+(7)/(11)sqrt(5b)=(7+sqrt(5))/(7-sqrt(5))-(7-sqrt(5))/(7+sqrt(5))=((7+sqrt(5))^(2)-(7-sqrt(5))^(2))/((7-sqrt(5))(7+sqrt(5)))`
`=((49+514sqrt(5))-(49+5-14sqrt(5)))/(49-25)`
`=(28sqrt(5))/(24)=0+(7)/(11)sqrt(5)`
Comparing we get `:." "a=0,b=1`