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Answer 1

$\mathsf{Given :\;x = \;\dfrac{2 - \sqrt{3}}{2 + \sqrt{3}}}$

$\mathsf{Multiplying\;the\;Numerator\;and\;Denominator\;with\;(2 - \sqrt{3}), We\;get :}$

$\mathsf{\implies x = \dfrac{(2 - \sqrt{3})(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})}}$

★  We know that : (a + b)(a - b) = a² - b²

$\mathsf{\implies x = \dfrac{(2 - \sqrt{3})^2}{(2)^2 - (\sqrt{3})^2}}$

★  We know that : (a - b)² = a² - 2ab + b²

$\mathsf{\implies x = \dfrac{(2)^2 + (\sqrt{3})^2 - (2)(2)\sqrt{3}}{4 - 3}}$

$\mathsf{\implies x = {4 + 3 - 4\sqrt{3}}}$

$\mathsf{\implies x = {7 - 4\sqrt{3}}}$

$\mathsf{\implies \dfrac{1}{x} = \dfrac{1}{7 - 4\sqrt{3}}}$

$\mathsf{Multiplying\;the\;Numerator\;and\;Denominator\;with\;(7 + 4\sqrt{3}), We\;get :}$

$\mathsf{\implies \dfrac{1}{x} = \dfrac{7 + 4\sqrt{3}}{(7 - 4\sqrt{3})(7 + 4\sqrt{3})}}$

$\mathsf{\implies \dfrac{1}{x} = \dfrac{7 + 4\sqrt{3}}{(7)^2 - (4\sqrt{3})^2}}$

$\mathsf{\implies \dfrac{1}{x} = \dfrac{7 + 4\sqrt{3}}{49 - (4)^2(\sqrt{3})^2}}$

$\mathsf{\implies \dfrac{1}{x} = \dfrac{7 + 4\sqrt{3}}{49 - (16)(3)}}$

$\mathsf{\implies \dfrac{1}{x} = \dfrac{7 + 4\sqrt{3}}{49 - 48}}$

$\mathsf{\implies \dfrac{1}{x} = {7 + 4\sqrt{3}}}$

$\mathsf{\implies x + \dfrac{1}{x} = (7 - 4\sqrt{3}) + ({7 + 4\sqrt{3})}}$

$\mathsf{\implies x + \dfrac{1}{x} = (7 + 7 - 4\sqrt{3} + 4\sqrt{3})}}$

$\mathsf{\implies x + \dfrac{1}{x} = (7 + 7)}}$

$\mathsf{\implies x + \dfrac{1}{x} = 14}}$

Answer 2

ANSWER

14

Step By step Explanation

Here we have =>

x= $\dfrac{2 - \sqrt 3}{2+\sqrt 3}$

=>x= $\dfrac{(2 - \sqrt 3)\times(2-\sqrt 3)}{(2+\sqrt 3)(2-\sqrt 3)}$

=>x= $\dfrac{(2 - \sqrt 3)^2}{(2)^2 - (\sqrt 3)^2}$

{By using

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

=>x= $\dfrac{4 + 3 - 4\sqrt 3} {4-3}$

=>x= $7 - 4\sqrt 3$

Now,

The value of$\dfrac{1}{x}$ will be >

=> $\dfrac{1}{7 - 4\sqrt 3}$

Now, on multiplying by $7 + 4\sqrt 3$ in numerator and denominator we get=>

$\dfrac{(1)(7 + 4\sqrt 3)}{(7 - 4\sqrt 3)(7 + 4\sqrt 3)}$

By using

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

=> $\dfrac{7 + 4\sqrt 3}{49 - 48}$

=> $\dfrac{7 + 4\sqrt 3}{1}$

=> ${7 + 4\sqrt 3}$

Hence,

x + $\dfrac{1}{x}$

=> ${7 + 4\sqrt 3}$ + $7 - 4\sqrt 3$

=> 14

Remember

1) (A+ B)² = A² + B² + 2 A* B

2) A² - B² =(A+B)(A-B)