Question

[nghiemminhphuong]

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

Giải hệ phương trình

Answer 1

[Phùng Minh Quân]

đk: \(x+2y\ge0\)

\(x+2y=\sqrt{\frac{x^2+4y^2}{2}}+\sqrt{\frac{\left(x+y\right)^2}{3}+y^2}\ge\sqrt{\frac{\left(x+2y\right)^2}{4}}+\sqrt{\frac{\left(x+2y\right)^2}{4}}=x+2y\)

\(\Rightarrow\)\(x=2y\)\(\Rightarrow\)\(x=3-y=3-\frac{x}{2}\)\(\Rightarrow\)\(\hept{\begin{cases}x=2\\y=\frac{x}{2}=1\end{cases}}\)