Đáp án:
Giải thích các bước giải:
$\frac{1-3x}{2x}$ -$\frac{3x-2}{2x-1}$ +$\frac{3x-2}{2x-4x^2}$
=$\frac{1-3x}{2x}$ -$\frac{3x-2}{2x-1}$ +$\frac{3x-2}{2x(1-2x)}$
=$\frac{1-3x}{2x}$ -$\frac{3x-2}{2x-1}$ -$\frac{3x-2}{2x(2x-1)}$
=$\frac{(1-3x)(2x-1)}{2x(2x-1)}$ -$\frac{(3x-2)2x}{2x(2x-1)}$ -$\frac{3x-2}{2x(2x-1)}$
=$\frac{2x-1-6x^2+3x}{2x(2x-1)}$ -$\frac{6x^2-4x}{2x(2x-1)}$ -$\frac{3x-2}{2x(2x-1)}$
=$\frac{(2x-1-6x^2+3x)-(6x^2-4x)-(3x-2)}{2x(2x-1)}$
=$\frac{2x-1-6x^2+3x-6x^2+4x-3x+2}{2x(2x-1)}$
=$\frac{-12x^2+6x+1}{2x(2x-1)}$
