HPT \(\Leftrightarrow\hept{\begin{cases}y-x=2\\\frac{x+y}{xy}=\frac{4}{3}\end{cases}\Leftrightarrow\hept{\begin{cases}x=y-2\\\frac{2y-2}{\left(y-2\right)y}=\frac{4}{3}\left(1\right)\end{cases}}}\)
Từ ( 1 ) suy ra \(3\left(2y-2\right)=4y\left(y-2\right)\)
\(\Rightarrow4y^2-14y+6=0\Rightarrow\orbr{\begin{cases}y=3\\y=\frac{1}{2}\end{cases}}\)
+) y = 3 suy ra x = 1
+) y = \(\frac{1}{2}\)suy ra x = \(\frac{-3}{2}\)
\(\hept{\begin{cases}\frac{y}{2}-\frac{x}{2}=1\\\frac{1}{x}+\frac{1}{y}=\frac{4}{3}\end{cases}\left(x,y>0\right)\Leftrightarrow\hept{\begin{cases}y-x=2\\\frac{y+x}{xy}=\frac{4}{3}\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}-x+y=2\\3.\left(x+y\right)=4.xy\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}-3x+3y=6\\3x+3y=4xy\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}-x+y=2\\6y=6+4xy\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=y-2\\6y=6+4.\left(y-2\right).y\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=y-2\\6y=6+4y^2-8y\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=y-2\\4y^2-14y+6=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=y-2\\\left(y-3\right).\left(4y-2\right)=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=y-2\\y=3;y=\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=1\left(TM\right);x=\frac{-3}{4}\left(L\right)\\y=3\left(TM\right);y=\frac{1}{2}\left(L\right)\end{cases}}\)
Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(1;3\right)\)
