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 : \(\left\{{}\begin{matrix}\dfrac{x}{x-1}=2\\\dfrac{1}{y+2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+2=1\\2x-2=x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=2\end{matrix}\right.\)
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