Question

√6/√3-√2, Rationalise the denominators of the given numbers.

Answer 1

$\bold{Answer :}$

We multiply both the numerator and the denominator of the given number by its conjugate irrational number (√3 + √2) of its denominator.

Then,

(√6)/(√3 - √2)

= {(√6)(√3 + √2)}/{√3 + √2)(√3 - √2)}

= (3√2 + 2√3)/(3 - 2)

= (3√2 + 2√3)/1,

which is the required form.

#$\bold{MarkAsBrainliest}$

Answer 2

$= \frac{\sqrt{6} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ = \frac{ \sqrt{18} + \sqrt{12} }{( \sqrt{3} )^{2} - ( { { \sqrt{2} } })^{2} } \\ = \frac{3 \sqrt{2} + 2 \sqrt{3} }{3 - 2 } \\ = 3 \sqrt{2} + 2 \sqrt{3}$