Chắc là \(\dfrac{1}{y+2}\) chứ lẻ 2 vậy chuyển sang = 3 ??- Đặt \(\left\{{}\begin{matrix}\dfrac{x}{x-1}=a\\\dfrac{1}{y+2}=b\end{matrix}\right.\)HPTTT : \(\left\{{}\begin{matrix}3a-2b=4\\2a+b=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3a-2b=4\\4a+2b=10\end{matrix}\right.\)\(\Rightarrow7a=14\)\(\Rightarrow a=2\)\(\Rightarrow b=1\)- Thay lại ta được : \(\left\{{}\begin{matrix}\dfrac{x}{x-1}=2\\\dfrac{1}{y+2}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y+2=1\\2x-2=x\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=2\end{matrix}\right.\)Vậy ...Câu trả lời: Chắc là \(\dfrac{1}{y+2}\) chứ lẻ 2 vậy chuyển sang = 3 ??- Đặt \(\left\{{}\begin{matrix}\dfrac{x}{x-1}=a\\\dfrac{1}{y+2}=b\end{matrix}\right.\)HPTTT : \(\left\{{}\begin{matrix}3a-2b=4\\2a+b=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3a-2b=4\\4a+2b=10\end{matrix}\right.\)\(\Rightarrow7a=14\)\(\Rightarrow a=2\)\(\Rightarrow b=1\)- Thay lại ta được : \(\left\{{}\begin{matrix}\dfrac{x}{x-1}=2\\\dfrac{1}{y+2}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y+2=1\\2x-2=x\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=2\end{matrix}\right.\)Vậy ...