a: Sửa đề: \(\begin{cases}\sqrt3\cdot x-y=1\\ 2x+\sqrt2\cdot y=\sqrt3\end{cases}\)
\(\Rightarrow\begin{cases}\sqrt6\cdot x-y\cdot\sqrt2=\sqrt2\\ 2x+y\cdot\sqrt2=\sqrt3\end{cases}\Rightarrow\begin{cases}\sqrt6\cdot x-y\cdot\sqrt2+2x+y\cdot\sqrt2=\sqrt2+\sqrt3\\ y=\sqrt3\cdot x-1\end{cases}\)
=>\(\begin{cases}x\left(2+\sqrt6\right)=\sqrt2+\sqrt3\\ y=x\sqrt3-1\end{cases}\Rightarrow\begin{cases}x=\frac{\sqrt2+\sqrt3}{2+\sqrt6}=\frac{\sqrt2+\sqrt3}{\sqrt2\left(\sqrt2+\sqrt3\right)}=\frac{1}{\sqrt2}=\frac{\sqrt2}{2}\\ y=x\sqrt3-1=\frac{\sqrt2}{2}\cdot\sqrt3-1=\frac{\sqrt6-2}{2}\end{cases}\)
b: \(\begin{cases}x\left(y-3\right)+y=xy-2\\ \left(x+2\right)^2=x^2-2y\end{cases}\Rightarrow\begin{cases}xy-3x+y=xy-2\\ x^2+4x+4=x^2-2y\end{cases}\)
=>\(\begin{cases}-3x+y=-2\\ 4x+4=-2y\end{cases}\Rightarrow\begin{cases}y=-2+3x\\ 4x+4=-2\left(3x-2\right)=-6x+4\end{cases}\)
=>\(\begin{cases}10x=0\\ y=3x-2\end{cases}\Rightarrow\begin{cases}x=0\\ y=3\cdot0-2=-2\end{cases}\)
