Đáp án: $(x,y)\in\{(1, -2), (10, 13)\}$
Giải thích các bước giải:
ĐKXĐ: $x\ne 2$
Ta có:
$\dfrac{x-1}{3}=\dfrac{y+2}{5}=\dfrac{x+y+1}{x-2}=\dfrac{x-1+y+2}{3+5}=\dfrac{x+y+1}{8}$
$\to \dfrac{x+y+1}{x-2}=\dfrac{x+y+1}{8}$
Nếu $x+y+1=0$
$\to \dfrac{x-1}{3}=\dfrac{y+2}{5}=0$
$\to x=1, y=-2$
Nếu $x+y+1\ne 0$
$\to x-2=8\to x=10$
$\to \dfrac{10-1}{3}=\dfrac{y+2}{5}$
$\to 3=\dfrac{y+2}{5}$
$\to y+2=15$
$\to y=13$
