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Home » B. $ \frace^-x(1 + e^-x)^2 $ - Abu Waleed Tea

B. $ \frace^-x(1 + e^-x)^2 $ - Abu Waleed Tea

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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We want to compute:
\frac{a^2 + b^2}{a^2 - b^2} + \frac{a^2 - b^2}{a^2 + b^2}
Note that:
a^2 + b^2 = b^2(x^2 + 1), \quad a^2 - b^2 = b^2(x^2 - 1)
So the expression becomes:
\frac{b^2(x^2 + 1)}{b^2(x^2 - 1)} + \frac{b^2(x^2 - 1)}{b^2(x^2 + 1)}...
Let $ y = \frac{x^2 + 1}{x^2 - 1} $, then the expression becomes:
y + \frac{1}{y}
We compute $ y + \frac{1}{y} $. First, recall that $ x + \frac{1}{x} =...
x^2 + 2 + \frac{1}{x^2} = 25 \Rightarrow x^2 + \frac{1}{x^2} = 23

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