\(\left\{{}\begin{matrix}\sqrt{2}x-\sqrt{3}y=1\left(1\right)\\x+\sqrt{3}y=\sqrt{2}\left(2\right)\end{matrix}\right.\)
Lấy \(\left(1\right)+\left(2\right):\)
\(\sqrt{2}x+x-\sqrt{3}y+\sqrt{3}y=1+\sqrt{2}\)
\(\Rightarrow\sqrt{2}x+x-\sqrt{2}-1=0\)
\(\Rightarrow x\left(1+\sqrt{2}\right)-\left(1+\sqrt{2}\right)=0\)
\(\Rightarrow\left(1+\sqrt{2}\right)\left(x-1\right)=0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
Thay \(x=1\) vào \(\left(2\right):1+\sqrt{3}y=\sqrt{2}\)
\(\Rightarrow\sqrt{3}y=\sqrt{2}-1\)
\(\Rightarrow y=\dfrac{\sqrt{2}-1}{\sqrt{3}}\)
Vậy hệ pt có nghiệm duy nhất \( \left(x;y\right)=\left(1;\dfrac{\sqrt{2}-1}{\sqrt{3}}\right)\)
\(\left\{{}\begin{matrix}\sqrt{2}x-\sqrt{3}y=1\\x+\sqrt{3}y=\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{2}+1\right)x=1+\sqrt{2}\\x+\sqrt{3}y=\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1+\sqrt{2}}{\sqrt{2}+1}=1\\x+\sqrt{3}y=\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\1+\sqrt{3}y=\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{\sqrt{2}-1}{\sqrt{3}}\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm duy nhất \(\left(x;y\right)=\left(1;\dfrac{\sqrt{2}-1}{\sqrt{3}}\right)\)
