If `x=sqrt6+sqrt5` then `x^2+1/x^2-2=`

Explanation

By Rationalizing denominator, we have :
`= 1 × (sqrt6 -sqrt5) / [(sqrt6 + sqrt5) × (sqrt6 - sqrt5)]`
`= (sqrt6 - sqrt5) / (sqrt(6))^2 - (sqrt(5))^2`
[We know that, `(a + b) (a - b) = a^2 - b^2`]
`= (sqrt 6 - sqrt5)/(6 - 5)`
`= (sqrt6 - sqrt5 )/1`
`= sqrt6 - sqrt5`
`1/x = sqrt6 - sqrt5` ………(2)
Then, `x^2 + 1/x^2 - 2`
We know that, `x^2 + 1/x^2 - 2 = (x - 1/x)^2`
On putting the value from eq 1 & 2 :
`x^2 + 1/x^2 - 2 = [sqrt6 + sqrt5 - (sqrt6 - sqrt5 )]^2`
`x^2 + 1/x^2 - 2 = [sqrt6 + sqrt5 - sqrt6 + sqrt5 ]^2`
`x^2 + 1/x^2 - 2 = (sqrt5 + sqrt5)^2`
`x^2 + 1/x^2 - 2 = (2sqrt5)^2`
`x^2 + 1/x^2 - 2 = 4 xx 5`
`x^2 + 1/x^2 - 2 = 20`