Question

\(\begin{cases}\sqrt{3}x-\sqrt{2}y=1\\\sqrt{2}x+\sqrt{3}=2\sqrt{6}\end{cases} \)

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Answer 1

Căn 3 y ạ

Answer 2

\(\Leftrightarrow\left\{{}\begin{matrix}3x-\sqrt{6}y=\sqrt{3}\\2x+\sqrt{6}y=4\sqrt{3}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5x=5\sqrt{3}\\2x+\sqrt{6}y=4\sqrt{3}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\sqrt{3}\\y=\frac{4\sqrt{3}-2x}{\sqrt{6}}=\sqrt{2}\end{matrix}\right.\)

Answer 3

\(\left\{{}\begin{matrix}\sqrt{3}x-\sqrt{2}y=1\\\sqrt{2}x+\sqrt{3}y=2\sqrt{6}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x-\sqrt{6}y=\sqrt{3}\\2x+\sqrt{6}y=4\sqrt{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x=5\sqrt{3}\\2x+\sqrt{6}y=4\sqrt{3}\end{matrix}\right.\)

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

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