Question

3+2√2/3-√2. rationalise the denominator

Answer 1

Identity used :

$(a + b)(a - b) = {a}^{2} - {b}^{2}$

$\frac{3 + 2 \sqrt{2} }{3 - \sqrt{2} }$

On rationalizing the denominator we get,

$= \frac{3 + 2 \sqrt{2} }{3 - \sqrt{2} } \times \frac{3 + \sqrt{2} }{3 + \sqrt{2} } \\ \\ = \frac{3(3 + \sqrt{2} ) + 2 \sqrt{2} (3 + \sqrt{2} )}{ {(3)}^{2} - {( \sqrt{2} )}^{2} } \\ \\ = \frac{ 9 + 3 \sqrt{2} + 6 \sqrt{2} + 4}{9 - 2} \\ \\ = \frac{13 + 9 \sqrt{2} }{7}$

Answer 2

$\textbf{Answer:}$

$\frac{3 + 2 \sqrt{2} }{3 - \sqrt{2} }$

$= \frac{3 + 2 \sqrt{2} }{3 - \sqrt{2} } \times \frac{3 + \sqrt{2} }{3 + \sqrt{2} }$

Now, we use the identity,

(a + b)(a - b) = a² - b²

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


$= \frac{9 + 3 \sqrt{2} + 6 \sqrt{2} + 4 }{9 - 2}$

$= \frac{13 + 9 \sqrt{2} }{7}$
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