Question

[Bao Cao Su]

giải hệ phương trình

\(\hept{\begin{cases}\sqrt{2x}+2\sqrt{3y}=5\\3\sqrt{2x}-\sqrt{3y}=\frac{9}{2}\end{cases}}\)

Answer 1

[Nguyễn Công Tỉnh]

\(\hept{\begin{cases}\sqrt{2x}+2\sqrt{3y}=5\\3\sqrt{2x}-\sqrt{3y}=\frac{9}{2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{2x}+2\sqrt{3y}=5\\6\sqrt{2x}-2\sqrt{3y}=9\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{2x}+2\sqrt{3y}=5\\7\sqrt{2x}=14\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{2x}+2\sqrt{3y}=5\\\sqrt{2x}=2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{2x}+2\sqrt{3y}=5\\2x=2^2=4\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{2.2}+2\sqrt{3y}=5\\x=2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2+2\sqrt{3y}=5\\x=2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2\sqrt{3y}=3\\x=2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{3y}=\frac{3}{2}\\x=2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}3y=\left(\frac{3}{2}\right)^2=\frac{9}{4}\\x=2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=\frac{3}{4}\\x=2\end{cases}}\)