Question

Evaluate (x-1/x)^2-(x+1/x)^2​

Answer 1

$\: \: \: {(x - \frac{1}{x} )}^{2} - {(x + \frac{1}{x}) }^{2} \\ \\ = ( {x}^{2} + \frac{1}{ {x}^{2} } - 2.x. \frac{1}{x} ) - ( {x}^{2} + \frac{1}{ {x}^{2} } + 2.x. \frac{1}{x} ) \\ \\ = {x}^{2} + \frac{1}{ {x}^{2} } - 2.x. \frac{1}{ {x} } - {x}^{2} - \frac{1}{ {x}^{2} } - 2.x. \frac{1}{x} \\ \\ = - 2 - 2 \\ \\ = - 4$

Answer 2

${\bold{\underline{\sf {\red{question}}}}}$

Evaluate :-

$\sf ( x - \dfrac{1}{x})^2 - ( x + \dfrac{1}{x})^2$

${\bold{\underline{\sf {\pink{Solution}}}}}$

This can be done by 2 methods

${\bold{\blue{\boxed{\bf{1st \:method}}}}}$

$\implies\sf ( x - \dfrac{1}{x})^2 - ( x + \dfrac{1}{x})^2$

${\bold{\blue{\boxed{\bf{a^2 - b^2 = (a+b)(a-b) }}}}}$

$\sf\implies(x- \dfrac{1}{x} + x + \dfrac{1}{x})(x- \dfrac{1}{x} - x -\dfrac{1}{x})$

$\sf\implies( x- \cancel\dfrac{1}{x} + x + \cancel\dfrac{1}{x})(\cancel x- \dfrac{1}{x} - \cancel x -\dfrac{1}{ x})$

$\sf\implies(2 x)(\dfrac{-2}{ x})$

$\sf\implies(2\cancel x)(\dfrac{-2}{ \cancel x})$

$\sf\implies - 4$

${\bold{\blue{\boxed{\bf{\dag -4}}}}}$

${\bold{\green{\boxed{\bf{2nd\: method}}}}}$

$\sf\implies \: \: \: {(x - \frac{1}{x} )}^{2} - {(x + \frac{1}{x}) }^{2}$

${\bold{\green{\boxed{\bf{(x+y)^2 = x^2 + y^2 + 2xy}}}}}$

${\bold{\green{\boxed{\bf{(x+y)^2 = x^2 + y^2 - 2xy}}}}}$

$\\ \\ \sf\implies ( {x}^{2} + \frac{1}{ {x}^{2} } - 2.x. \frac{1}{x} ) - ( {x}^{2} + \frac{1}{ {x}^{2} } + 2.x. \frac{1}{x} ) \\ \\ \sf\implies {x}^{2} + \frac{1}{ {x}^{2} } - 2.x. \frac{1}{ { x} } - {x}^{2} - \frac{1}{ {x}^{2} } - 2.x. \frac{1}{x} \\ \\ \sf\implies - 2 - 2 \\ \\ \sf\implies - 4$

${\bold{\green{\boxed{\bf{\dag - 4}}}}}$

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