Question

evalute if x=1/√3+√2 find x+1/x

Answer 1

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Here is the answer you were looking for:
$x = \frac{1}{ \sqrt{3} + \sqrt{2} } \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ x = \frac{1}{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ by \: using \: the \: identity \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ \sqrt{3} - \sqrt{2} }{ {( \sqrt{3}) }^{2} - {( \sqrt{2}) }^{2} } \\ \\ = \frac{ \sqrt{3} - \sqrt{2} }{3 - 2} \\ \\ x = \sqrt{3} - \sqrt{2} \\ \\ \frac{1}{x} = \frac{1}{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ = \frac{ \sqrt{3} + \sqrt{2} }{ {( \sqrt{3} )}^{2} - {( \sqrt{2} )}^{2} } \\ \\ = \frac{ \sqrt{3} + \sqrt{2} }{3 - 2} \\ \\ \frac{1}{x} = \sqrt{3} + \sqrt{2} \\ \\ x + \frac{1}{x} = ( \sqrt{3} - \sqrt{2} ) + ( \sqrt{3} + \sqrt{2} ) \\ \\ x + \frac{1}{x} = \sqrt{3} - \sqrt{2} + \sqrt{3} + \sqrt{2} \\ \\ x + \frac{1}{x} = \sqrt{3} + \sqrt{3} \\ \\ x + \frac{1}{x} = 2 \sqrt{3}$

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