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\(a-b=2\left(a+b\right)\)
\(\Rightarrow a-b=2a+2b\)
\(\Rightarrow a=-3b\)
\(a-b=a.b\)
\(\Rightarrow-3b-b=\left(-3b\right).b\)
\(\Rightarrow-4b=-3b^2\)
\(\Rightarrow3b^2-4b=0\Rightarrow b\left(3b-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}b=0\\b=\dfrac{4}{3}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=0\\b=0\end{matrix}\right.\\\left\{{}\begin{matrix}a=-4\\b=\dfrac{4}{3}\end{matrix}\right.\end{matrix}\right.\)
\(a-b=2\left(a+b\right)\\ \Leftrightarrow a-b=2a+2b\\ \Leftrightarrow a=-3b\\ a-b=ab\Leftrightarrow-4b=-3b^2\Leftrightarrow3b^2-4b=0\\ \Leftrightarrow\left[{}\begin{matrix}b=\dfrac{4}{3}\\b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=-4\\b=0\end{matrix}\right.\)
Vậy \(\left(a;b\right)=\left\{\left(0;0\right);\left(-4;\dfrac{4}{3}\right)\right\}\)
