Question

Tìm giá trị lớn nhất của \(A=\sqrt{5x-4}+\sqrt{12-5x}\)

Answer 1

Áp dụng bđt:\(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\) ta có:

\(A=\sqrt{5x-4}+\sqrt{12-5x}\ge\sqrt{5x-4+12-5x}=\sqrt{8}\)

Dấu "=" xảy ra khi:\(\left(5x-4\right)\left(12-5x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=\dfrac{12}{5}\end{matrix}\right.\)

Answer 2

Áp dụng BĐT Bunhiacopxki

\(\Rightarrow A=1.\sqrt{5x-4}+1.\sqrt{12-5x}\le\sqrt{\left(1^2+1^2\right)\left(5x-4+12-5x\right)}\)

\(\Rightarrow A\le\sqrt{2.8}=4\)

vậy MaxA=4 khi \(5x-4=12-5x\Leftrightarrow10x=16\Leftrightarrow x=\dfrac{8}{5}\)