Решите уравнение: [latex]\frac{1-3x}{1-2x}+\frac{5}{(2x-1)^{2}}-3=0[/latex]

[latex]\frac{1-3x}{1-2x}+\frac{5}{(2x-1)^2}-3=0;\\\\ 2x-1 \neq 0; x \neq 0.5;\\\\ \frac{(1-3x)(1-2x)}{(1-2x)^2}+\frac{5}{(1-2x)^2}-\frac{3(1-2x)^2}{(1-2x)^2}=0;\\\\ (1-3x)(1-2x)+5-3(1-2x)^2=0;\\\\ 1-3x-2x+6x^2+5-3+12x-12x^2=0;\\\\ -6x^2+7x+3=0;\\\\ 6x^2-7x-3=0;\\\\ D=(-7)^2-4*6*(-3)=121=11^2;\\\\ x_1=\frac{7+11}{2*6}=1.5;\\\\ x_2=\frac{7-11}{2*6}=-\frac{1}{3};[/latex]