Question

root6+root3by root6-root3 rationalize the denominator

Answer 1

hiii...

please make as brainalist

Answer 2

Hi there !!

Given,
to rationalize the denominator in:


$\frac{ \sqrt{6} + \sqrt{3} }{ \sqrt{6} - \sqrt{ 3} }$
Using the rationalising identity in the denominator ie, a²-b²=(a-b)(a+b),
we have,


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

In the numerator, using the identity (a+b)²= a²+2ab+b² and simplyfying the denominator,
we have,

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

$= \frac{6 + 2 \sqrt{18} + 3 }{3}$

2✓18 can be written as 6✓2 since 2✓18 = 2✓2×3×3 = 2×3✓2 = 6✓2

So,

$\frac{6 + 6 \sqrt{2} + 3 }{3} = \frac{9 + 6 \sqrt{2} }{3}$
Taking 3 as common factor in the numerator,
we have,

$= \frac{3(3 + 2 \sqrt{2}) }{3}$

cancelling out 3 in numerator and denominator,
we are left with

3 + 2✓2
is the answer

___________________________________

Hope helped