Question

plzzzzzzzz solve it it's urgent

Answer 1

Heya there !!!
Here is the answer you were looking for:

Identity used :

$(x + y)(x - y) = {x}^{2} - {y}^{2}$

$\frac{4}{3 \sqrt{3} - 2 \sqrt{2} } + \frac{3}{3 \sqrt{3} + 2 \sqrt{2} }$

On rationalizing the denominators we get,

$\frac{4}{3 \sqrt{3} - 2 \sqrt{2} } \times \frac{3 \sqrt{3} + 2 \sqrt{2} }{3 \sqrt{3} + 2 \sqrt{2} } + \frac{3}{3 \sqrt{3} + 2 \sqrt{2} } \times \frac{3 \sqrt{3} - 2 \sqrt{2} }{3 \sqrt{3} - 2 \sqrt{2} } \\ \\ = \frac{4(3 \sqrt{3} + 2 \sqrt{2} )}{ {(3 \sqrt{3}) }^{2} - {(2 \sqrt{2}) }^{2} } + \frac{3(3 \sqrt{3} - 2 \sqrt{2}) }{ {(3 \sqrt{3} )}^{2} - {(2 \sqrt{2} )}^{2} } \\ \\ = \frac{12 \sqrt{3} + 8 \sqrt{2} }{27 - 8} + \frac{9 \sqrt{3} - 6 \sqrt{2} }{27 - 8} \\ \\ = \frac{12 \sqrt{3} + 8 \sqrt{2} }{19} + \frac{9 \sqrt{3} - 6 \sqrt{2} }{19} \\ \\ = \frac{12 \sqrt{3} + 8 \sqrt{2} + 9 \sqrt{3} - 6 \sqrt{2} }{19} \\ \\ = \frac{21 \sqrt{3} + 2 \sqrt{2} }{19} \\ \\ = \frac{21 \times 1.732 + 2 \times 1.414}{19} \\ \\ = \frac{36.372 + 2.828}{19} \\ \\ = \frac{39.2}{19} \\ \\ = 2.06 \: (approx)$

Hope this helps!!!

If you have any doubt regarding to this answer, feel free to ask me in the comment section ^_^

@Mahak24

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Answer 2

Hello here ...

Solution here
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we have
$\frac{4}{3 \sqrt{2} - 2 \sqrt{2} } + \frac{3}{3 \sqrt{3} + 2 \sqrt{2} }$

$= > \frac{4(3 \sqrt{3} + 2 \sqrt{2} ) + 3(3 \sqrt{3} - 2 \sqrt{2} ) }{(3 \sqrt{3} - 2 \sqrt{2} )(3 \sqrt{3} + 2 \sqrt{2} )} \\ \\ = > \frac{12 \sqrt{3} + 8 \sqrt{2} + 9 \sqrt{3} - 6 \sqrt{2} }{(3 \sqrt{3} {)}^{2} - (2 \sqrt{2} {)}^{2} }$
[using formula (a + b ) ( a - b)=a^2 - b^2]

$= > \frac{21 \sqrt{3} + 2 \sqrt{2} }{27 - 8} = \frac{21 \sqrt{3} + 2 \sqrt{2} }{19} \\ \\ = > \frac{21 \times 1.732 + 2 \times 1.414}{19} \\ \\ = > \frac{36.372 + 2.828}{19} = \frac{39.2}{19} \\ \\ = > 2.06316 = 2.063. \: \: answer$
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