1/ Ta có \(\left(x-2\right)\left(x+\frac{2}{3}\right)>0\)
=> \(\hept{\begin{cases}x-2>0\\x+\frac{2}{3}>0\end{cases}}\)hoặc \(\hept{\begin{cases}x-2< 0\\x+\frac{2}{3}< 0\end{cases}}\)
=> \(\hept{\begin{cases}x>2\\x>-\frac{2}{3}\end{cases}}\)hoặc \(\hept{\begin{cases}x< 2\\x< -\frac{2}{3}\end{cases}}\)
=> \(\orbr{\begin{cases}x>2\\x< -\frac{2}{3}\end{cases}}\)
Vậy \(\orbr{\begin{cases}x>2\\x< -\frac{2}{3}\end{cases}}\)thì \(\left(x-2\right)\left(x+\frac{2}{3}\right)>0\)
2 \(xy=\frac{x}{y}\Rightarrow y=\frac{x}{xy}=\frac{1}{y}\Rightarrow y^2=1\Rightarrow y=+-1\)
nếu \(y=1\Rightarrow x+y=xy=x+1=x\Rightarrow x-x=-1\Rightarrow0=-1\)vô lí (loại)
\(\Rightarrow y=-1\Rightarrow x+y=xy=x-1=-x\Rightarrow2x=1\Rightarrow x=\frac{1}{2}\)(thỏa mãn)
vậy \(x=\frac{1}{2};y=-1\)