If , then Find the Value of

1.	If , then Find the Value of
Answer» Solution!! X = 1 - √2 To FIND :- (x - 1/x)² $\sf \left(1-\sqrt{2}-\dfrac{1}{1-\sqrt{2}}\right)^{2}$ Rationalise the denominator. $\sf = \left(1-\sqrt{2}-\left(\dfrac{1}{1-\sqrt{2}}\times \dfrac{1+\sqrt{2}}{1+\sqrt{2}}\right)\right)^{2}$ $\sf = \left(1-\sqrt{2}-\left(\dfrac{1(1+\sqrt{2})}{(1-\sqrt{2})(1+\sqrt{2})}\right)\right)^{2}$ Use the identity → (a - B)(a + b) = a² - b². $\sf = \left(1-\sqrt{2}-\left(\dfrac{1+\sqrt{2}}{(1)^{2}-(\sqrt{2})^{2}}\right)\right)^{2}$ $\sf = \left(1-\sqrt{2}-\left(\dfrac{1+\sqrt{2}}{1-2}\right)\right)^{2}$ $\sf = \left(1-\sqrt{2}-\left(\dfrac{1+\sqrt{2}}{-1}\right)\right)^{2}$ $\sf = \left(1-\sqrt{2}-\left(-\left(1+\sqrt{2}\right)\right)\right)^{2}$ Opening the brackets and CHANGING the sign. $\sf = \left(1-\sqrt{2}+\left(1+\sqrt{2}\right)\right)^{2}$ $\sf = \left(1-\sqrt{2}+1+\sqrt{2}\right)^{2}$ Grouping the like term. $\sf = \left(1+1-\sqrt{2}+\sqrt{2}\right)^{2}$ $\sf = \left(1+1\right)^{2}$ $\sf = \left(2\right)^{2}$ $\sf = 4$

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