Câu 1a)
\(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y-2}=4\\\dfrac{4}{x}-\dfrac{1}{y-2}=1\end{matrix}\right.\left(I\right)\)
Đặt: \(\left\{{}\begin{matrix}a=\dfrac{2}{x}\left(x\ne0\right)\\b=\dfrac{1}{y-2}\left(y\ne2\right)\end{matrix}\right.\)
\(\left(I\right)\Leftrightarrow\left\{{}\begin{matrix}a+3b=4\\2a-b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}a=\dfrac{2}{x}=1\\b=\dfrac{1}{y-2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\left(TM\right)\\y=3\left(TM\right)\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm (x;y)=(2;3)
Câu 1:
a) Ta có: \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y-2}=4\\\dfrac{4}{x}-\dfrac{1}{y-2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x}+\dfrac{6}{y-2}=8\\\dfrac{4}{x}-\dfrac{1}{y-2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{7}{y-2}=7\\\dfrac{4}{x}-\dfrac{1}{y-2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y-2=1\\\dfrac{4}{x}-1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\\dfrac{4}{x}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là (x,y)=(2;3)
