Question

\(\left\{{}\begin{matrix}\frac{1}{x}+\frac{3}{y-2}=4\\\frac{4}{x}+\frac{1}{y-2}=1\end{matrix}\right.\)

Answer 1

\(\frac{1}{x}\) sửa thành \(\frac{2}{x}\)nha tìm đk rồi giải hpt hộ mình

Answer 2

Điều kiện xác định: \(\left\{{}\begin{matrix}x\ne0\\y\ne2\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\frac{2}{x}+\frac{3}{y-2}=4\\\frac{4}{x}+\frac{1}{y-2}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{4}{x}+\frac{6}{y-2}=8\\\frac{4}{x}+\frac{1}{y-2}=1\end{matrix}\right.\Leftrightarrow\left(\frac{4}{x}+\frac{6}{y-2}\right)-\left(\frac{4}{x}+\frac{1}{y-2}\right)=8-1\)

\(\Leftrightarrow\frac{5}{y-2}=7\Leftrightarrow y=\frac{19}{7}\)\(\Rightarrow x=-10\)

Vậy...........