Решите дробное рациональное уравнение

Решите задачу:

$\mathtt{\frac{3x^2+11x-4}{3x-1}=3;~\frac{3(x+4)(x-\frac{1}{3})}{3(x-\frac{1}{3})}=3,~\to~x=-1}$

$\mathtt{\frac{7}{x-3}+1=\frac{18}{x^2-6x+9};~1=\frac{18}{(x-3)^2}-\frac{7}{x-3}=\frac{18-7(x-3)}{(x-3)^2};~}\\\mathtt{(x-3)^2+7(x-3)-18=0;\left[\begin{array}{ccc}\mathtt{x-3=-9}\\\mathtt{x-3=2}\end{array}\right\to~\left[\begin{array}{ccc}\mathtt{x=-6}\\\mathtt{x=5}\end{array}\right}$

$\mathtt{\frac{2x^2+x-1}{2x-1}=2;~\frac{2(x+1)(x-\frac{1}{2})}{2(x-\frac{1}{2})}=2,~\to~x=1}$

$\mathtt{\frac{5}{x-2}+1=\frac{14}{x^2-4x+4};~1=\frac{14}{(x-2)^2}-\frac{5}{x-2}=\frac{14-5(x-2)}{(x-2)^2};~}\\\mathtt{(x-2)^2+5(x-2)-14=0;\left[\begin{array}{ccc}\mathtt{x-2=-7}\\\mathtt{x-2=2}\end{array}\right\to~\left[\begin{array}{ccc}\mathtt{x=-5}\\\mathtt{x=4}\end{array}\right}$