If \(\frac{1+\sqrt{2}}{1-\sqrt{2}}\) is expressed in the form of x+y√2 find the values of x and y
\(\frac{1+\sqrt{2}}{1-\sqrt{2}} \times \frac{1+\sqrt{2}}{1+\sqrt{2}}\\
=\frac{1+(1+\sqrt{2})+\sqrt{2}(1+\sqrt{2})}{1^2 - (\sqrt{2})^2}\\
=\frac{(1+\sqrt{2}+\sqrt{2}+2)}{1-2}\\
=\frac{3+2\sqrt{2}}{-1}\\
=-3-2\sqrt{2}\\
∴X and Y = -3 and -2\)
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