Question

Answer 1

\(\left\{{}\begin{matrix}\left(\sqrt{3}-\sqrt{2}\right)x+y=\sqrt{2}\\x+\left(\sqrt{3}+\sqrt{2}\right)y=\sqrt{6}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)x+\left(\sqrt{3}+\sqrt{2}\right)y=\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)\\x+\left(\sqrt{3}+\sqrt{2}\right)y=\sqrt{6}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x+y\left(\sqrt{3}+\sqrt{2}\right)=\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)\\x+y\left(\sqrt{3}+\sqrt{2}\right)=\sqrt{6}\end{matrix}\right.\)

Vì \(\dfrac{1}{1}=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\ne\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{6}}\)

nên hệ phương trình vô nghiệm