Question

${\boxed {\LARGE{\mathfrak {solve\: it:-}}}}$

$\tt \left (\dfrac {x+1}{x-1}\right)^2-\left (\dfrac {x+1}{x-1}\right)=2$​

Answer 1

$\tt \left (\dfrac {x+1}{x-1}\right)^2-\left (\dfrac {x+1}{x-1}\right)=2$

Answer

$\frac{x + 1}{x - 1} {}^{2} - \frac{x + 1}{x - 1} = 2$

Answer 2

$\large \pmb{\sf{Question :}}$

$\tt \left (\dfrac {x+1}{x-1}\right)^2-\left (\dfrac {x+1}{x-1}\right)=2$

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$\large{\pmb{\sf{Final \: Solution:}}}$

$\red \bigstar\color{green} \boxed{ \sf \purple{x = 0}} \\ \red \bigstar\color{green} \boxed{ \sf \purple{x = 3}}$

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$\large{\pmb{\sf{Step-by-step \; Explanation :}}}$

$\sf{ :\leadsto\left (\dfrac {x+1}{x-1}\right)^2-\left (\dfrac {x+1}{x-1}\right)=2}$

$\sf{ :\leadsto\left (\dfrac {x+1}{x-1}\right)^2-\left (\dfrac {x+1}{x-1}\right) - 2 = 0}$

$\footnotesize \boxed{ \green{ \sf{Let \: \orange t = \left(\dfrac {x+1}{x-1}\right)}}}$

$\sf{ :\leadsto {t}^{2} - t - 2 = 0}$

$\sf\therefore t = -1 \: \: ; \: \: t = 2$

$\footnotesize \boxed{ \green{ \sf{Substituting \: back \: \orange t = \left(\dfrac {x+1}{x-1}\right)}}}$

$\sf :\leadsto \left(\dfrac {x+1}{x-1}\right) = - 1\: \: ; \: \:\left(\dfrac {x+1}{x-1}\right) = 2$

$\pink \bigstar \: \color{maroon} \boxed{ \color{green}\sf :\leadsto x = 3\: \: ; \: \: x = 0}$

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$\sf \color{azure}\fcolorbox{pink}{black}{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: @Saanvigrover2007 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:}$