Câu 4.Hệ $\Leftrightarrow \begin{cases}2x^2-\frac{2}{y^2}-(\sqrt{2}+1)x\sqrt{2}-\frac{2x}{2x^2+\frac{2}{y^2}}=-(1+\sqrt{2}) \\ \frac{4x}{y}-(\sqrt{2}+1)\frac{\sqrt{2}}{y}+\frac{\frac{2}{y}}{2x^2+\frac{2}{y^2}}=0 \end{cases}$ $(1)$
Đặt $\begin{cases}a=x\sqrt{2} \\ b=\frac{\sqrt{2}}{y}\neq 0\end{cases}$ thì $(1)\Leftrightarrow \begin{cases}a^2-b^2-(\sqrt{2}+1)a-\frac{a\sqrt{2}}{a^2+b^2}=-(1+\sqrt{2}) (2)\\ 2ab-(\sqrt{2}+1)b+\frac{b\sqrt{2}}{a^2+b^2}=0 (3)\end{cases}$
Lấy $(1)+i.(2)\Rightarrow (a^2-b^2+2ab)-(\sqrt{2}+1)(a+bi)-\frac{\sqrt{2}(a-bi)}{a^2+b^2}+\sqrt{2}+1=0(4)$
Đặt $z=a+bi$ thì $(4)\Leftrightarrow z^2-(\sqrt{2}+1)z-\frac{\sqrt{2}(a-bi)}{a^2+b^2}+\sqrt{2}+1=0$
$\Leftrightarrow (z-\sqrt{2})(z^2-z+1)=0$
$\Rightarrow x=\frac{\sqrt{2}}{4}\wedge y=\frac{2\sqrt{6}}{3}$ hoặc $x=\frac{\sqrt{2}}{4}\wedge y=-\frac{2\sqrt{6}}{3}$