$\left\{\begin{matrix}(1+\sqrt{2})x+(1-\sqrt{2})y=5 & \\ (1+\sqrt{2})x+(1+\sqrt{2})y=3 & \end{matrix}\right.$
Trừ phương trình thứ nhất cho phương trình thứ hai ta có hệ:
$\left\{\begin{matrix}[1-\sqrt{2}-(1+\sqrt{2})]y=2 & \\ (1+\sqrt{2})x+(1+\sqrt{2})y=3 & \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}(1-\sqrt{2}-1-\sqrt{2})y=2 & \\ (1+\sqrt{2})x+(1+\sqrt{2})y=3 & \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}-2\sqrt{2}y=2 & \\ (1+\sqrt{2})x+(1+\sqrt{2})y=3 & \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}y=\frac{2}{-2\sqrt{2}} & \\ (1+\sqrt{2})x+(1+\sqrt{2})y=3 & \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}y=\frac{-1}{\sqrt{2}} & \\ (1+\sqrt{2})x+(1+\sqrt{2})y=3 & \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}y=\frac{-1}{\sqrt{2}} & \\ (1+\sqrt{2})x+(1+\sqrt{2})\frac{-1}{\sqrt{2}}=3 & \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}y=\frac{-1}{\sqrt{2}} & \\ (1+\sqrt{2})x-\frac{1+\sqrt{2}}{\sqrt{2}}=3 & \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}y=\frac{-1}{\sqrt{2}} & \\ (1+\sqrt{2})x=3+\frac{1+\sqrt{2}}{\sqrt{2}} & \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}y=\frac{-1}{\sqrt{2}} & \\ (1+\sqrt{2})x=\frac{3\sqrt{2}}{\sqrt{2}}+\frac{1+\sqrt{2}}{\sqrt{2}} & \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}y=\frac{-1}{\sqrt{2}} & \\ (1+\sqrt{2})x=\frac{3\sqrt{2}+1+\sqrt{2}}{\sqrt{2}} & \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}y=\frac{-1}{\sqrt{2}} & \\ (1+\sqrt{2})x=\frac{4\sqrt{2}+1}{\sqrt{2}} & \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}y=\frac{-1}{\sqrt{2}} & \\ x=\frac{4\sqrt{2}+1}{\sqrt{2}}\div (1+\sqrt{2}) & \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}y=\frac{-1}{\sqrt{2}} & \\ x=\frac{4\sqrt{2}+1}{\sqrt{2}.(1+\sqrt{2})} & \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}y=\frac{-1}{\sqrt{2}} & \\ x=\frac{4\sqrt{2}+1}{\sqrt{2}+2} & \end{matrix}\right.$
Vậy hệ phương trình có một nghiệm duy nhất là $\left ( \frac{4\sqrt{2}+1}{\sqrt{2}+2};\frac{-1}{\sqrt{2}} \right )$
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