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Problem 641

AIME late

Algebra Difficulty 5.0 Find the answer

Solve the inequality $\frac{2 x^{2}-10 x+11}{x^{2}-6 x+8} \leqslant 1$ .

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Official solution

Analysis The most commonly used method for solving rational inequalities is the factorization method.
Solution Transform the original inequality to $\frac{2 x^{2}-10 x+11}{x^{2}-6 x+8}-1 \leqslant 0$ .
Rearranging and simplifying, we get $\frac{(x-1)(x-3)}{(x-2)(x-4)} \leqslant 0$ . That is, $\left\{\begin{array}{l}(x-1)(x-2)(x-3)(x-4) \leqslant 0 \\ x \neq 2, x \neq 4\end{array}\right.$ .
From this, we can conclude that the solution set of the original inequality is: $\{x \mid 1 \leqslant x0(<0)$ , we only need to find the intervals where $f(x)$ is positive or negative.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.