Question

if x=1÷(2-√3) then the value of x-1÷x

Answer 1

Hey friend,
Here is the answer you were looking for:
$x = \frac{1}{2 - \sqrt{3} } \\ \\ on \: rationalizing \: we \: get \\ \\ x = \frac{1}{2 - \sqrt{3} } \times \frac{2 + \sqrt{3} }{2 + \sqrt{3} } \\ \\ using \: the \: identity \\ (a + b)(a - b) = { a }^{2} - {b}^{2} \\ \\ = \frac{2 + \sqrt{3} }{ {(2)}^{2} - { (\sqrt{3} )}^{2} } \\ \\ = \frac{2 + \sqrt{3} }{4 - 3} \\ \\ x = 2 + \sqrt{3} \\ \\ \frac{1}{x} = \frac{1}{2 + \sqrt{3} } \\ \\ on \: rationalizing \: we \: get \\ \\ \frac{1}{x} = \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \\ using \: same \: identity \\ \\ \frac{1}{x} = \frac{2 - \sqrt{3} }{ {(2)}^{2} - {( \sqrt{3}) }^{2} } \\ \\ = \frac{2 - \sqrt{3} }{4 - \sqrt{3} } \\ \\ \frac{1}{x} = 2 - \sqrt{3} \\ \\ x - \frac{1}{x} = (2 + \sqrt{3} ) - (2 - \sqrt{3} ) \\ \\ x - \frac{1}{x} = 2 + \sqrt{3} - 2 + \sqrt{3} \\ \\ x - \frac{1}{x} = \sqrt{3} + \sqrt{3} \\ \\ x - \frac{1}{x} = 2 \sqrt{3}$

Hope this helps!!!!

@Mahak24

Thanks...
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