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Home » We are given $ f(1) = 1 $, so: - IX Labs

We are given $ f(1) = 1 $, so: - IX Labs

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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Thus, the number **with** at least one pair of consecutive 3s is:
- 55 = \boxed{201}
\frac{a + 2b}{a - 2b} + \frac{a - 2b}{a + 2b} = 2.
Find the value of $ \frac{a^2 + 4b^2}{a^2 - 4b^2} $.
Let’s simplify the left-hand side. Let $ x = \frac{a + 2b}{a - 2b} $....
x + \frac{1}{x} = 2 \Rightarrow x^2 - 2x + 1 = 0 \Rightarrow (x - 1)^2...
\frac{a + 2b}{a - 2b} = 1.
Cross-multiplying:
a + 2b = a - 2b \Rightarrow 2b = -2b \Rightarrow 4b = 0 \Rightarrow b...
Now substitute $ b = 0 $ into the expression we are to evaluate:

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