`1/(sqrt(9)- sqrt(8))` is equal to:
(a)`3+2sqrt(2)`
(b) `1/(3+2sqrt(2))`
(c) `3-2sqrt(2)`
(d) `3/2- sqrt(2)`

Explanation

`1/(sqrt9-sqrt8)=1/(3-2sqrt2)`
[∵`sqrt8 =sqrt(2xx2xx2)=2sqrt2`]
`1/(3-2sqrt2)xx(3+2sqrt2)/(3+2sqrt2)`
[multiplying numerator and denominator by `3+ 2sqrt2`]
=`(3+2sqrt2)/(9-(2sqrt2)^2`
[using identity `(a - b)(a + b) = a^2 - b^2]`
`=(3+2sqrt2)/(9-8)`
`=(3+2sqrt2)`