Question

please answer this question........ ​

Answer 1

Answer:

7+'^2\left(7+4\sqrt{3}\right)-2'\:-4\sqrt{3}

Step-by-step explanation:

Answer 2

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$here \: given \: that : \\ x = \frac{1}{2 - \sqrt{3} } \\ = > x = \frac{1 \times (2 + \sqrt{3}) }{(2 - \sqrt{3})(2 + \sqrt{3} )} \\ (multyply \frac{2 + \sqrt{3} }{2 + \sqrt{3} } = 1) \\ = > x = \frac{2 + \sqrt{3} }{ {2}^{2} - { \sqrt{3} }^{2} } \\ (becaue \: of \: {(x + y)}^{2} = (x + y)(x + y) \\ here \: we \: can \: compair \: 2 \: as \: x \: \\ and \: \sqrt{3} = y) \\ = > x = \frac{2 + \sqrt{3} }{4 - 3} = 2 + \sqrt{3} \\ now \: we \: have \: to \: find \: the \: value \: of \\ {x}^{2} =( {2 + \sqrt{3}) }^{2} \\ = > {x}^{2} = {(2)}^{2} + 2.2. \sqrt{3} + {( \sqrt{3} )}^{2} \\ we \: know \: that \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ = > {x}^{2} = 4 + 4 \sqrt{3} + 3 \\ = > {x}^{2} = 7 + 4 \sqrt{3} \\ according \: to \: the \: question \\ {(x - \frac{1}{x} )}^{2} \\ = ( { \frac{ {x}^{2} - 1}{x} )}^{2} \\ we \: found \: that \\ x = 2 + \sqrt{3} \\ and \: {x}^{2} = 7 + 4 \sqrt{3} \\ now \: putting \: the \: values \\ ( { \frac{7 + 4 \sqrt{3} - 1}{2 + \sqrt{3} }) }^{2} \\ = ( { \frac{6 + 4 \sqrt{3} }{2 + \sqrt{3} }) }^{2} \\ = ( { \frac{2(3 + 2 \sqrt{3}) }{2 + \sqrt{3} }) }^{2} \\ = ( { \frac{2 \times \sqrt{3} ( \sqrt{3 } + 2) }{2 + \sqrt{3} } )}^{2} \\ we \: know \: that \: \sqrt{3} \times \sqrt{3} = 3) \\ = ( { \frac{2 \sqrt{3} ( \sqrt{3 } + 2) }{2 + \sqrt{3} }) }^{2} \\ =( {2 \sqrt{3} )}^{2} \\ = 2 \times 2 \times \sqrt{3} \times \sqrt{3} \\ = 4 \times 3 \\ = 12$

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the answer is 12 ✔✔✔✔✔

✨✨hope it helps✨✨