Question

please send fast today is my test.

Answer 1

HEY UR ANSWER IS HERE....

Answer 2

Hey Friend,
Here is the answer you were looking for:
$\frac{22}{2 \sqrt{3} + 1} + \frac{17}{2 \sqrt{3} - 1} \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ = \frac{22}{2 \sqrt{3} + 1} \times \frac{2 \sqrt{3} - 1 }{2 \sqrt{3} - 1} + \frac{17}{2 \sqrt{3} - 1} \times \frac{2 \sqrt{3} + 1}{2 \sqrt{3} + 1 } \\ \\ using \: the \: identity \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{22 \times 2 \sqrt{3} - 22 \times 1 }{ {(2 \sqrt{3}) }^{2} - {(1)}^{2} } + \frac{17 \times 2 \sqrt{3} + 17 \times 1 }{ {(2 \sqrt{3} )}^{2} - {(1)}^{2} } \\ \\ = \frac{44 \sqrt{3} - 22 }{12 - 1} + \frac{3 \sqrt{3} + 17 }{12 - 1} \\ \\ = \frac{44 \sqrt{3} - 22 }{11} + \frac{3 \sqrt{3} + 17 }{11} \\ \\ = 4 \sqrt{3} - 2 + \frac{3 \sqrt{3} + 17 }{11}$

Hope this helps!!!!

@Mahak24

Thanks....
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