Rozwiążmy równania podane w ramce:

$x-\frac{2x-1}{3}=0\mid \cdot 3$  

$3x-\left(2x-1\right)=0$  

$3x-2x+1=0$    

$x+1=0$  

$x=-1\in (-3;1⟩$  


$\frac{2x}{1-2x}=\frac{x+1}{2-x}$  

założenia:

$1-2x\ne 0\text{i}2-x\ne 0$  

$-2x\ne -1\text{i}-x\ne -2$  

$x\ne \frac{1}{2}\text{i}x\ne 2$  

zatem:

$2x\cdot \left(2-x\right)=\left(x+1\right)\left(1-2x\right)$  

$4x-2x^2=x-2x^2+1-2x\mid +2x^2$  

$4x=x+1-2x$  

$5x=1\mid :5$  

$x=\frac{1}{5}\in (-3;1⟩$  


$8x-3\left(2x-1\right)=6$  

$8x-6x+3=6$  

$2x=3$  

$x=\frac{3}{2}\notin (-3;1⟩$  


$\left(3x-2\right)\left(3x+2\right)=3x\left(3x+2\right)$  

$9x^2-4=9x^2+6x$  

$6x=-4$  

$x=-\frac{4}{6}$  

$x=-\frac{2}{3}\in (-3;1⟩$  


Odp.: C