Решите неравенство

$\mathtt{|\frac{x-5}{3x^2-5x-2}|\geq1~\to\left[\begin{array}{ccc}\mathtt{\frac{x-5}{3x^2-5x-2}+1\leq0}\\\mathtt{\frac{x-5}{3x^2-5x-2}-1\geq0}\end{array}\right\to~\left[\begin{array}{ccc}\mathtt{\frac{x-5+(3x^2-5x-2)}{3x^2-5x-2}\leq0}\\\mathtt{\frac{x-5-(3x^2-5x-2)}{3x^2-5x-2}\geq0}\end{array}\right\to~\\}$ $\\\mathtt{\left[\begin{array}{ccc}\mathtt{\frac{3x^2-4x-7}{3x^2-5x-2}\leq0}\\\mathtt{\frac{3x^2-6x+3}{3x^2-5x-2}\leq0}\end{array}\right\to~\left[\begin{array}{ccc}\mathtt{\frac{(x+1)(x-\frac{7}{3})}{(x+\frac{1}{3})(x-2)}\leq0}\\\mathtt{\frac{(x-1)^2}{(x+\frac{1}{3})(x-2)}\leq0}\end{array}\right\to~\left[\begin{array}{ccc}\mathtt{x\in[-1;-\frac{1}{3})U(2;\frac{7}{3}]}\\\mathtt{x\in(-\frac{1}{3};2)}\end{array}\right}$

ответ: $\mathtt{x\in[-1;-\frac{1}{3})U(-\frac{1}{3};2)U(2;\frac{7}{3}]}$