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