Решение. Задание 15, Тренировочная работа 24.01.2019. Вариант Восток

Условие задачи

Решите неравенство $\displaystyle {{{log}_2 \left(x+1\right)}}^2\cdot {{log}_{\frac{1}{3}} x^2}-4{{log}_2 \left(x+1\right)}+4{{log}_3 \left(-x\right)}+4\leq 0.$

Решение

$\displaystyle {{{log}_2 \left(x+1\right)}}^2\cdot {{log}_{\frac{1}{3}} x^2}-4{{log}_2 \left(x+1\right)}+4{{log}_3 \left(-x\right)}+4\leq 0.$

Найдём ОДЗ: $\left\{ \begin{array}{c}x+1\textgreater 0, \\-x\textgreater 0 \end{array}\right.\Leftrightarrow \left\{ \begin{array}{c}x\textgreater -1, \\x\textless 0 \end{array}\right.\Leftrightarrow -1\textless x\textless 0.$

Упростим ${{{log}_2 \left(x+1\right)}}^2=2{{log}_2 \left|x+1\right|}=2{{log}_2 \left(x+1\right)},$ при $x+1\textgreater 0.$

$\displaystyle {{log}_{\frac{1}{3}} x^2}=2{{log}_{\frac{1}{3}} \left|x\right|}=2{{log}_{\frac{1}{3}} \left(-x\right)},$ при $-x\textgreater 0.$ Кроме того,

$\displaystyle 2{{log}_{\frac{1}{3}} \left(-x\right)}=2\cdot \frac{{{log}_3 \left(-x\right)}}{{{log}_3 \left(\frac{1}{3}\right)}}=-2{{log}_3 \left(-x\right)}$ и переписываем неравенство

$-2\cdot 2{{log}_2 \left(x+1\right)}\cdot {{log}_3 \left(-x\right)}-4{{log}_2 \left(x+1\right)}+4{{log}_3 \left(-x\right)}+4\leq 0.$ Сделаем замену

${{log}_2 \left(x+1\right)}=t, \; {{log}_3 \left(-x\right)}=z,$ тогда

$\displaystyle -4tz-4t+4z+4\le 0\left.\right|\cdot \left(-\frac{1}{4}\right), \; tz+t-z-1\geq 0,$

$t\left(z+1\right)-\left(z+1\right)\geq 0, \; \left(z+1\right)\left(t-1\right)\geq 0.$

Вернёмся к переменной x.

$\left\{ \begin{array}{c}\left({{log}_3 \left(-x\right)}+1\right)\left({{log}_2 \left(x+1\right)}-1\right)\geq 0, \\-1\textless x\textless 0. \end{array}\right.$

Записывая $-1={{log}_3 3^{-1}}, \ {{log}_3 \left(-x\right)}+1={{log}_3 \left(-x\right)}-{{log}_3 3^{-1}}$ и используя метод замены множителя, получаем

$\left\{ \begin{array}{c}\left(-x-\frac{1}{3}\right)\left(x-1\right)\geq 0, \\-1\textless x\textless 0 \end{array}\right.\Leftrightarrow \left\{ \begin{array}{c}\left(x+\frac{1}{3}\right)\left(x-1\right)\leq 0, \\-1\textless x\textless 0. \end{array}\right.$