Question

Cho 2 số thực x, y thỏa mãn:

\(\left(x+\sqrt{x^2+2020}\right)\left(2y+\sqrt{4y^2+2020}\right)=2020\)

Tìm GTLN cuẩ biểu thức: B=\(\dfrac{x^2}{2}+4xy+3y^2+x+3y+15\)

Answer 1

\(\left(x+\sqrt{x^2+2020}\right)\left(2y+\sqrt{\left(2y\right)^2+2020}\right)=2020\)

\(\Leftrightarrow\left\{{}\begin{matrix}2y+\sqrt{\left(2y\right)^2+2020}=\sqrt{x^2+2020}-x\\x+\sqrt{x^2+2020}=\sqrt{\left(2y\right)^2+2020}-2y\end{matrix}\right.\)

\(\Rightarrow x+2y+\sqrt{x^2+2020}+\sqrt{\left(2y\right)^2+2020}=-x-2y+\sqrt{x^2+2020}+\sqrt{\left(2y\right)^2+2020}\)

\(\Leftrightarrow2\left(x+2y\right)=0\)

\(\Leftrightarrow x=-2y\)

\(\Rightarrow B=2y^2-8y^2+3y^2-2y+3y+15\)

\(\Rightarrow B=-3y^2+y+15=-3\left(y-\dfrac{1}{6}\right)^2+\dfrac{181}{12}\)

\(B_{max}=\dfrac{181}{12}\) khi \(y=\dfrac{1}{6}\)