Đáp án:
$(x;y)=(3;2)$
Giải thích các bước giải:
$\begin{cases}\dfrac{3}{x-2}+\dfrac{2}{y+1}=\dfrac{11}{3}\\\dfrac{2x-2}{x-2}+\dfrac{y}{y+1}=\dfrac{14}{3}\end{cases}$
$\begin{cases}\dfrac{3}{x-2}+\dfrac{2}{y+1}=\dfrac{11}{3}\\\dfrac{2x-4+2}{x-2}+\dfrac{y+1-1}{y+1}=\dfrac{14}{3}\end{cases}$
$\begin{cases}\dfrac{3}{x-2}+\dfrac{2}{y+1}=\dfrac{11}{3}\\\dfrac{2x-4+2}{x-2}+\dfrac{y+1-1}{y+1}=\dfrac{14}{3}\end{cases}$
$\begin{cases}\dfrac{3}{x-2}+\dfrac{2}{y+1}=\dfrac{11}{3}\\\dfrac{2(x-2)+2}{x-2}+\dfrac{(y+1)-1}{y+1}=\dfrac{14}{3}\end{cases}$
$\begin{cases}\dfrac{3}{x-2}+\dfrac{2}{y+1}=\dfrac{11}{3}\\2+\dfrac{2}{x-2}+1+\dfrac{-1}{y+1}=\dfrac{14}{3}\end{cases}$
Đặt $\dfrac{1}{x-2}=u;\dfrac{1}{y+1}=v$ ta có :
$\begin{cases}3u+2y=\dfrac{11}{3}\\2+2u+1-v=\dfrac{14}{3}\end{cases}$
$\begin{cases}3u+2y=\dfrac{11}{3}\\2u-v=\dfrac{5}{3}\end{cases}$
$\begin{cases}3u+2y=\dfrac{11}{3}\\4u-2v=\dfrac{10}{3}\end{cases}$
$\begin{cases}4u=\dfrac{21}{3}\\4u-2v=\dfrac{10}{3}\end{cases}$
$\begin{cases}u=1\\4.1-2v=\dfrac{10}{3}\end{cases}$
$\begin{cases}u=1\\-2v=\dfrac{-2}{3}\end{cases}$
$\begin{cases}u=1\\v=\dfrac{1}{3}\end{cases}$
Với $u=1$ thì :
$x-2=1$
$x=3$
Với $v=\dfrac{1}{3}$
$y+1=3$
$y=2$
Vậy hệ pt có nghiệm duy nhất :
$(x;y)=(3;2)$
