Question

 

Решите уравнение:

$frac{1-3x}{1-2x}+frac{5}{(2x-1)^{2}}-3=0$

Answer 1

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