Prove that `3sqrt(48) -4 sqrt(75) + sqrt(192) = 0`

1.	Prove that `3sqrt(48) -4 sqrt(75) + sqrt(192) = 0`
Answer» `3 sqrt(48) -4 sqrt(75) + sqrt(192) = 3 sqrt(16 xx 3) -4 sqrt(25 xx 3) + sqrt(64 xx 3)` `= 3 xx 4 sqrt3 -4 xx 5 sqrt3 + 8sqrt3` `= 12 sqrt3 - 20 sqrt3 + 8 sqrt3` `= 20 sqrt3 - 20 sqrt3 = 0` Hence, `3 sqrt(48) - 4 sqrt(75) + sqrt(192) = 0`

Discussion

If `x = 2 + sqrt3, y = 2 - sqrt3`, then find the simpliest value of (a) `x - (1)/(x)`, (b) `y^(2) + (1)/(y^(2))`, (c) `x^(3) - (1)/(x^(3))`, (d) `xy + (1)/(xy)`
If `m + (1)/(m) = sqrt3`, then find the simpliest value of (i) `m^(2) + (1)/(m^(2))` and (ii) `m^(3) + (1)/(m^(3))`:
The simpliest value of `(sqrt5 (sqrt(3 - sqrt5)))/(sqrt2 + sqrt(7 - 3 sqrt5))` isA. 1B. 5C. 10D. None of these
Rationalise the denominator: (a) `(1)/(root(3)(3) + root(3)(2))`, (b) `(2)/(sqrt5 + sqrt3 + sqrt2)`, (c) `(x^(2))/(sqrt(x^(2) + y^(2)) - y)`, (d) `(1)/(sqrt6 + sqrt5 - sqrt11)` (e) `(sqrt(x + 2y) - sqrt(x -2y))/(sqrt(x + 2y) + sqrt(x - 2y))`, (f) `(sqrt10 + sqrt5 - sqrt3)/(sqrt10 - sqrt5 + sqrt3)`
Prove that `(12)/(3 + sqrt5 - 2 sqrt2) = 1 + sqrt5 + sqrt10 - sqrt2`
`sqrt7 - sqrt5` is a binomial surd
Which of `(sqrt15 + sqrt3) and (sqrt10 + sqrt8)` is greater ?
Between `(sqrt6 + sqrt7) and (sqrt5 + sqrt8)`, the greater one is-A. `sqrt5 + sqrt8`B. `sqrt6 + sqrt7`C. `sqrt5 + 2 sqrt2`D. None of these
Which one is greater between `(sqrt7 - sqrt5) and (sqrt13 - sqrt11)` ?
Express in the form of like surds: `root(3)(4), root(6)(5), 2sqrt3`.