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Тренинги по решению вариантов ЕГЭ — 2020. Вариант 2. Задание 15. Решение

Условие задачи

Решить неравенство \displaystyle \frac{2}{5^{x+1}-1}+\frac{5^{x+1}-2}{5^{x+1}-3} \geq 2

Решение

Замена 5^{x+1}=z,z \, \textgreater \, 0

\displaystyle \frac{2}{z-1}+\frac{z-2}{z-3} \geq 2

\displaystyle \frac{2}{z-1}+\frac{z-2}{z-3}-2 \geq 0

\displaystyle \frac{2(z-3)+(z-2)(z-1)-2(z-1)(z-3)}{(z-1)(z-3)} \geq 0

\displaystyle \frac{-z^2+7z-10}{(z-1)(z-3)} \geq 0

\displaystyle \frac{z^2-7z+10}{(z-1)(z-3)} \leq 0

\displaystyle \frac{(z-2)(z-5)}{(z-1)(z-3)} \leq 0 \Leftrightarrow \left[ \begin{gathered} 1\, \textless z \leq 2 \\ 3\, \textless \, z\leq 5 \\ \end{gathered} \right. \Leftrightarrow \left[ \begin{gathered} 1\, \textless \,5^{x+1} \leq 2 \\ 3\, \textless\, 5^{x+1} \leq 5 \\ \end{gathered} \right.\Leftrightarrow \left[ \begin{gathered} 5^0\, \textless \, 5^{x+1} \leq 5^{log_ 5 2} \\ 5^{log_5 3} \, \textless \, 5^{x+1} \leq 5^1 \\ \end{gathered} \right. \Leftrightarrow

\displaystyle \left[ \begin{gathered} 0 \, \textless \, x+1\leq log_5 2 \\ log_ 5 3\, \textless \, x+1 \leq 1 \\ \end{gathered} \right. \Leftrightarrow \left[ \begin{gathered} -1 \, \textless \, x\leq log_ 2 \frac{2}{5} \\ log_5 \frac{3}{5}\, \textless \, x \leq0 \\ \end{gathered} \right.

Ответ

\displaystyle x \in (-1; log_5 \frac{2}{5} ] \cup (log_5 \frac{3}{5}; 0]