26.If \( \frac{1+\sqrt{2}}{1-\sqrt{2}}+\frac{1-\sqrt{2}}{1+\sqrt{2}}=a+b \sqrt{2} \), Find the value of \( a \) and \( b \)

To simplify the expression \( \frac{1+\sqrt{2}}{1-\sqrt{2}}+\frac{1-\sqrt{2}}{1+\sqrt{2}} \), we can combine the fractions over a common denominator. The common denominator is \( (1 - \sqrt{2})(1 + \sqrt{2}) = 1^2 - (\sqrt{2})^2 = 1 - 2 = -1 \). Now we rewrite each fraction: 1. \( \frac{1+\sqrt{2}}{1-\sqrt{2}} = \frac{(1+\sqrt{2})(1+\sqrt{2})}{(1-\sqrt{2})(1+\sqrt{2})} = \frac{(1+\sqrt{2})^2}{-1} = -((1+\sqrt{2})^2) \) Calculating \( (1+\sqrt{2})^2 \): \[ (1+\sqrt{2})^2 = 1 + 2 + 2\sqrt{2} = 3 + 2\sqrt{2} \] Thus, \[ \frac{1+\sqrt{2}}{1-\sqrt{2}} = -(3 + 2\sqrt{2}) = -3 - 2\sqrt{2} \] 2. Now for \( \frac{1-\sqrt{2}}{1+\sqrt{2}} \) in a similar fashion: \[ \frac{1-\sqrt{2}}{1+\sqrt{2}} = \frac{(1-\sqrt{2})(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})} = \frac{(1-\sqrt{2})^2}{-1} = -((1-\sqrt{2})^2) \] Calculating \( (1-\sqrt{2})^2 \): \[ (1-\sqrt{2})^2 = 1 - 2 + 2\sqrt{2} = -1 + 2\sqrt{2} \] Thus, \[ \frac{1-\sqrt{2}}{1+\sqrt{2}} = -(-1 + 2\sqrt{2}) = 1 - 2\sqrt{2} \] Now we combine both results: \[ \frac{1+\sqrt{2}}{1-\sqrt{2}} + \frac{1-\sqrt{2}}{1+\sqrt{2}} = (-3 - 2\sqrt{2}) + (1 - 2\sqrt{2}) \] Combine the constants and the terms with \(\sqrt{2}\): \[ = -3 + 1 - 2\sqrt{2} - 2\sqrt{2} = -2 - 4\sqrt{2} \] Now, matching the given format \( a + b\sqrt{2} \): \[ a = -2, \quad b = -4 \] Thus, the values of \( a \) and \( b \) are: \[ \boxed{-2} \quad \text{and} \quad \boxed{-4} \]