Помогите, пожалуйста, решить неравенство!!! [latex]log _{3x-5} (2 x^{2} -9x+10) \geq 0[/latex]

ОДЗ: [latex]\left\{\begin{array}{rcl}3x-5\ \textgreater \ 0 \\3x-5\neq1 \\2x^2-9x+10\ \textgreater \ 0\end{array}\right. , \left\{\begin{array}{rcl}x\ \textgreater \ \frac{5}{3} \\x\neq2 \\(x-\frac{5}{2})(x-1)\ \textgreater \ 0\end{array}\right., x\ \textgreater \ \frac{5}{2}[/latex] Решение: [latex]log_{3x-5}(2x^2-9x+10) \geq log_{3x-5}1\\1) \left \{ {{3x-5\ \textgreater \ 1} \atop {2x^2-9x+10\geq1}} \right. , \left \{ {{x\ \textgreater \ 2} \atop {2(x-\frac{3}{2})(x-3)\geq0}} \right. , \left \{ {{x\ \textgreater \ 2} \atop {x \leq \frac{3}{2}, x \geq 3}} \right. , x \geq 3[/latex] [latex]2) \left \{ {{3x-5\ \textless \ 1} \atop {2x^2-9x+10 \leq 1}} \right. ,\left \{ {{x\ \textless \ 2} \atop {(x-\frac{3}{2})(x-3) \leq 0}} \right. ,\left \{ {{x\ \textless \ 2} \atop {\frac{3}{2}\leq x\leq 3}} \right. ,\frac{3}{2}\leq x\ \textless \ 2[/latex] Не удовлетворяет ОДЗ. Ответ:[latex]x \geq 3[/latex]