Question

1. Cho x=\(\dfrac{1}{2}\sqrt{\sqrt{2}+\dfrac{1}{8}}-\dfrac{1}{8}\sqrt{2}\)

Tính giá trị của A= \(x^2+\sqrt{x^4+x+1}\)

2.Tính GTLN của: P=\(\dfrac{\sqrt{x-2018}}{x+2}+\dfrac{\sqrt{x-2019}}{x}\)

Answer 1

2.

\(P=\dfrac{\sqrt{x-2018}}{x+2}+\dfrac{\sqrt{x-2019}}{x}\)

\(P=\dfrac{\sqrt{\left(x-2018\right).2020}}{\left(x+2\right)\sqrt{2020}}+\dfrac{\sqrt{\left(x-2019\right).2019}}{\sqrt{2019}.x}\)

Áp dụng BĐT AM-GM:

\(\sqrt{\left(x-2018\right).2020}\le\dfrac{1}{2}\left(x-2018+2020\right)=\dfrac{1}{2}\left(x+2\right)\)

\(\sqrt{\left(x-2019\right).2019}\le\dfrac{1}{2}\left(x-2019+2019\right)=\dfrac{1}{2}x\)

\(\Rightarrow P\le\dfrac{x+2}{2\sqrt{2020}\left(x+2\right)}+\dfrac{x}{2\sqrt{2019}.x}=\dfrac{1}{2\sqrt{2020}}+\dfrac{1}{2\sqrt{2019}}\)

\("="\Leftrightarrow x=4038\)