Question

Tìm max :

A = 2x - x² + 2\(\sqrt{x^2-2x+3}\)

Answer 1

Đặt \(\sqrt{x^2-2x+3}=\sqrt{\left(x-1\right)^2+2}=t\Rightarrow\left\{{}\begin{matrix}t\ge\sqrt{2}\\2x-x^2=3-t^2\end{matrix}\right.\)

\(A=-t^2+2t+3=-t^2+2t+2-2\sqrt{2}+1+2\sqrt{2}\)

\(A=\left(\sqrt{2}-t\right)\left(\sqrt{2}+t\right)+2\left(t-\sqrt{2}\right)+1+2\sqrt{2}\)

\(A=\left(\sqrt{2}-t\right)\left(t+\sqrt{2}-2\right)+1+2\sqrt{2}\)

Do \(t\ge\sqrt{2}\Rightarrow\left\{{}\begin{matrix}\sqrt{2}-t\le0\\t+\sqrt{2}-2\ge2\sqrt{2}-2>0\end{matrix}\right.\)

\(\Rightarrow\left(\sqrt{2}-t\right)\left(t+\sqrt{2}-2\right)\le0\)

\(\Rightarrow A\le1+2\sqrt{2}\Rightarrow A_{max}=1+2\sqrt{2}\) khi \(t=\sqrt{2}\) hay \(x=1\)