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Nguyễn Lê Phước Thịnh · Accepted Answer

$\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-2+2}{x-2}+\dfrac{1}{y+1}=3\\dfrac{4}{x-2}-\dfrac{3}{y+1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x-2}+\dfrac{1}{y+1}=2\\dfrac{4}{x-2}-\dfrac{3}{y+1}=1\end{matrix}\right.$$\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{x-2}+\dfrac{3}{y+1}=6\\dfrac{4}{x-2}-\dfrac{3}{y+1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{10}{x-2}=7\\dfrac{2}{x-2}+\dfrac{1}{y+1}=2\end{matrix}\right.$$\Rightarrow\left\{{}\begin{matrix}x-2=\dfrac{10}{7}\\dfrac{1}{y+1}=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{24}{7}\y=\dfrac{2}{3}\end{matrix}\right.$Câu trả lời: $\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-2+2}{x-2}+\dfrac{1}{y+1}=3\\dfrac{4}{x-2}-\dfrac{3}{y+1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x-2}+\dfrac{1}{y+1}=2\\dfrac{4}{x-2}-\dfrac{3}{y+1}=1\end{matrix}\right.$$\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{x-2}+\dfrac{3}{y+1}=6\\dfrac{4}{x-2}-\dfrac{3}{y+1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{10}{x-2}=7\\dfrac{2}{x-2}+\dfrac{1}{y+1}=2\end{matrix}\right.$$\Rightarrow\left\{{}\begin{matrix}x-2=\dfrac{10}{7}\\dfrac{1}{y+1}=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{24}{7}\y=\dfrac{2}{3}\end{matrix}\right.$

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