Question

Simplify the following by rationalising the denominator.
Plz answer this question plz plz. ​

Answer 1

Answer:

1/3√2 - 2√3

By rationalising...

1/3√2-2√3 * 3√2+2√3/ 3√2+2√3

(a+b) (a-b) = a^2 - b^2

3√2-2√3/14. is the ans for first ques..

srry idk how to solve the sec one!

Answer 2

Step-by-step explanation:

Question:-

Solve the following by rationalizing the denominator:-

$\sf(i) \: \dfrac{1}{3 \sqrt{2} - 2 \sqrt{3} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (ii) \: \dfrac{3 \sqrt{5} - \sqrt{7} }{3 \sqrt{3} + \sqrt{2} }$

Solution:-

$\sf(i) \: \: \dfrac{1}{3 \sqrt{2} - 2 \sqrt{3} }$

By rationalizing the denominator:-

$\sf = \dfrac{1}{3 \sqrt{2} - 2 \sqrt{3} } \times \dfrac{3 \sqrt{2} + 2\sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3}}$

$\sf = \dfrac{3 \sqrt{2} + 2\sqrt{3} }{(3 \sqrt{2})^{2} - (2 \sqrt{3})^{2} }$

$\sf = \dfrac{3 \sqrt{2} + 2\sqrt{3} }{ }$

$\sf = \dfrac{3 \sqrt{2} + 2\sqrt{3} }{6 }$

$\underline{\boxed{\bf \leadsto \dfrac{3 \sqrt{2} + 2\sqrt{3} }{6 }}}$

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$(ii) \: \dfrac{3 \sqrt{5} - \sqrt{7} }{3 \sqrt{3} + \sqrt{2} }$

By rationalizing the denominator:-

$\sf = \dfrac{3 \sqrt{5} - \sqrt{7} }{3 \sqrt{3} + \sqrt{2} } \times \dfrac{3 \sqrt{3} - \sqrt{2} }{3 \sqrt{3} - \sqrt{2} }$

$\sf = \dfrac{3 \sqrt{5}( 3 \sqrt{3} - \sqrt{2}) - \sqrt{7}(3 \sqrt{3} - \sqrt{2}) }{(3 \sqrt{3} - \sqrt{2} )(3 \sqrt{3} + \sqrt{2}) }$

$\sf = \dfrac{3 \sqrt{5}( 3 \sqrt{3}) - 3 \sqrt{5} (\sqrt{2}) - \sqrt{7}(3 \sqrt{3} ) - \sqrt{7}( \sqrt{2}) }{(3 \sqrt{3})^{2} - (\sqrt{2} )^{2} }$

$\sf = \dfrac{9 \sqrt{15}- 3 \sqrt{10} - 3 \sqrt{21} - \sqrt{14} }{27 - 2 }$

$\sf = \dfrac{9 \sqrt{15}- 3 \sqrt{10} - 3 \sqrt{21} - \sqrt{14} }{25 }$

$\underline{\boxed{\bf \leadsto \dfrac{9 \sqrt{15}- 3 \sqrt{10} - 3 \sqrt{21} - \sqrt{14} }{25 } }}$