Giải thích các bước giải:

`a)C` xác định:

`<=>`$\begin{cases}2x-x \ne 0\\2x+x \ne 0\\x-y \ne 0 \end{cases}$

`<=>`$\begin{cases}x \ne 0\\x \ne y\end{cases}$

`b)``C=((x-2)/(2x-x)+3/(2x-x)-(x+3)/(2x+x)):(1-(x-3)/(x-y))`

`=((x-2)/x +3/x -(x+3)/(3x)):((x-y)/(x-y)-(x-3)/(x-y))`

`=((3(x-2))/(3x)+(3.3)/(3x)-(x+3)/(3x)):(x-y-x+3)/(x-y)`

`=(3x-6+9-x-3)/(3x) : (3-y)/(x-y)`

`=(2x)/(3x) . (x-y)/(3-y)`

`=(2(x-y))/(3(3-y))`

`=(2x-2y)/(9-3y)`

$a,$ $C$ xác định khi:

$2x-x\neq0→x\neq0$

$2x+x \neq 0→x\neq0$

$x-y \neq0→x\neq y $

 $b,$ $C=(\dfrac{x-2}{2x-x}+\dfrac{3}{2x-x}-\dfrac{x+3}{2x+x}):(1-\dfrac{x-3}{x-y})$ 

$C=(\dfrac{x-2}{x}+\dfrac{3}{x}-\dfrac{x+3}{3x}):(\dfrac{x-y-x+3}{x-y})$ 

$C=(\dfrac{3x-6}{3x}+\dfrac{9}{3x}-\dfrac{x+3}{3x}):(\dfrac{3-y}{x-y})$ 

$C=(\dfrac{3x-6+9-x-3}{3x}):(\dfrac{3-y}{x-y})$ 

$C=\dfrac{2x}{3x}:(\dfrac{3-y}{x-y})$ 

$C=\dfrac{2}{3}.\dfrac{x-y}{3-y})$ 

$C=\dfrac{2x-2y}{9x-3y}$