Question

Помогите, пожалуйста, решить неравенство!!!

$log _{3x-5} (2 x^{2} -9x+10) geq 0$

Answer 1

ОДЗ:
$left{begin{array}{rcl}3x-5 textgreater 0 \3x-5neq1 \2x^2-9x+10 textgreater 0end{array}right. , left{begin{array}{rcl}x textgreater frac{5}{3} \xneq2 \(x-frac{5}{2})(x-1) textgreater 0end{array}right., x textgreater frac{5}{2}$

Решение:
$log_{3x-5}(2x^2-9x+10) geq log_{3x-5}1\1) left { {{3x-5 textgreater 1} atop {2x^2-9x+10geq1}} right. , left { {{x textgreater 2} atop {2(x-frac{3}{2})(x-3)geq0}} right. , left { {{x textgreater 2} atop {x leq frac{3}{2}, x geq 3}} right. , x geq 3$
$2) left { {{3x-5 textless 1} atop {2x^2-9x+10 leq 1}} right. ,left { {{x textless 2} atop {(x-frac{3}{2})(x-3) leq 0}} right. ,left { {{x textless 2} atop {frac{3}{2}leq xleq 3}} right. ,frac{3}{2}leq x textless 2$
Не удовлетворяет ОДЗ.

Ответ:$x geq 3$