Giải hệ phương trình \(\left\{{}\begin{matrix}\left(\sqrt{3}-1\right)x-y=\sqrt{3}\\x+\left(\sqrt{3}+1\right)y=1\end{matrix}\right.\) bằng phương pháp thế, ta được nghiệm là
\(\left(x,y\right)=\left(\dfrac{4+\sqrt{3}}{3};-\dfrac{1}{3}\right)\).\(\left(x,y\right)=\left(\dfrac{4-\sqrt{3}}{3};-\dfrac{\sqrt{3}}{3}\right)\).\(\left(x,y\right)=\left(\dfrac{1-\sqrt{3}}{3};-\dfrac{1+\sqrt{3}}{3}\right)\).\(\left(x,y\right)=\left(\dfrac{4-2\sqrt{3}}{3};\dfrac{\sqrt{3}}{3}\right)\).Hướng dẫn giải:\(\left\{{}\begin{matrix}\left(\sqrt{3}-1\right)x-y=\sqrt{3}\\x+\left(\sqrt{3}+1\right)y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\left(\sqrt{3}-1\right)x-\sqrt{3}\\x+\left(\sqrt{3}+1\right)\left[\left(\sqrt{3}-1\right)x-\sqrt{3}\right]=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\left(\sqrt{3}-1\right)x-\sqrt{3}\\x+2x-\sqrt{3}\left(\sqrt{3}+1\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\left(\sqrt{3}-1\right)x-\sqrt{3}\\3x=4+\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\left(\sqrt{3}-1\right)\left(4+\sqrt{3}\right)}{3}-\sqrt{3}\\x=\dfrac{4+\sqrt{3}}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4+\sqrt{3}}{3}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
