If \( a = \sqrt{2} + 1 \) and \( b = \sqrt{2} - 1 \), then the value of \( \frac{1}{a+1} + \frac{1}{b+1} \) will be:
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Given \( a = \sqrt{2} + 1 \) and \( b = \sqrt{2} - 1 \), we need to find the value of \( \frac{1}{a+1} + \frac{1}{b+1} \).
Let's substitute the values of a and b into the expression:
\( \frac{1}{a+1} + \frac{1}{b+1} = \frac{1}{\sqrt{2} + 1 + 1} + \frac{1}{\sqrt{2} - 1 + 1} \)
\( = \frac{1}{\sqrt{2} + 2} + \frac{1}{\sqrt{2}} \)
To simplify the first term, we can rationalize the denominator:
\( \frac{1}{\sqrt{2} + 2} = \frac{1}{\sqrt{2} + 2} \times \frac{\sqrt{2} - 2}{\sqrt{2} - 2} = \frac{\sqrt{2} - 2}{2 - 4} = \frac{\sqrt{2} - 2}{-2} = \frac{2 - \sqrt{2}}{2} \)
Now, substitute this back into the expression:
\( \frac{2 - \sqrt{2}}{2} + \frac{1}{\sqrt{2}} = \frac{2 - \sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{2} + \sqrt{2}}{2} = \frac{2}{2} = 1 \)
Therefore, the value of \( \frac{1}{a+1} + \frac{1}{b+1} \) is 1.
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