Задание №75607 ЕГЭ Профильной математике

С помощью замены $3^{\mathit{x}}=\mathit{t}$, где $\mathit{t}>0$ приведём неравенство к виду

$\frac{35\mathit{t}}{4+10\mathit{t}-6{\mathit{t}}^2}\ge \frac{\mathit{t}+2}{3\mathit{t}+1}-\frac{3\mathit{t}-1}{\mathit{t}-2}$.

$-6{\mathit{t}}^2+10\mathit{t}+4=-2\left(3{\mathit{t}}^2-5\mathit{t}-2\right)=-2\left(\mathit{t}-2\right)\left(3\mathit{t}+1\right)$.

$\frac{35\mathit{t}}{-2\left(\mathit{t}-2\right)\left(3\mathit{t}+1\right)}\ge \frac{\left(\mathit{t}+2\right)\left(\mathit{t}-2\right)-\left(3\mathit{t}-1\right)\left(3\mathit{t}+1\right)}{\left(3\mathit{t}+1\right)\left(\mathit{t}-2\right)}$

$\frac{35\mathit{t}}{-2\left(\mathit{t}-2\right)\left(3\mathit{t}+1\right)}\ge \frac{\left({\mathit{t}}^2-4\right)-\left(9{\mathit{t}}^2-1\right)}{\left(3\mathit{t}+1\right)\left(\mathit{t}-2\right)}$

$\frac{35\mathit{t}}{-2\left(\mathit{t}-2\right)\left(3\mathit{t}+1\right)}\ge \frac{-8{\mathit{t}}^2-3}{\left(3\mathit{t}+1\right)\left(\mathit{t}-2\right)};$

$\frac{35\mathit{t}}{\left(\mathit{t}-2\right)\left(3\mathit{t}+1\right)}\le \frac{16{\mathit{t}}^2+6}{\left(3\mathit{t}+1\right)\left(\mathit{t}-2\right)};$

$\frac{16{\mathit{t}}^2-35\mathit{t}+6}{\left(3\mathit{t}+1\right)\left(\mathit{t}-2\right)}\ge 0;$

$\frac{16\left(\mathit{t}-2\right)\left(\mathit{t}-\frac{3}{16}\right)}{\left(3\mathit{t}+1\right)\left(\mathit{t}-2\right)}\ge 0;$

$\frac{\left(\mathit{t}-\frac{3}{16}\right)}{\left(3\mathit{t}+1\right)}\ge 0,\mathit{t}\ne 2.$

 или . С учётом условия . Возвращаясь к переменной $\mathit{x}$, получим, что  или $3^{\mathit{x}}>2$, откуда  или $\mathit{x}>\mathit{l}\mathit{o}{\mathit{g}}_{3}2$.