Question

Tìm gtln của A =\(\sqrt{13-x}+\sqrt{x-5}\)

Answer 1

\(A^2=\left(13-x\right)+\left(x-5\right)+2\sqrt{\left(13-x\right)\left(x-5\right)}\)

\(=8+2\sqrt{\left(13-x\right)\left(x-5\right)}\)(Dùng Bđt Cauchy)

\(\le8+\left(13-x\right)+\left(x-5\right)\)

\(=8+8=16\)

\(\Rightarrow A^2\le16\Leftrightarrow A\le4\)

Dấu = khi \(\sqrt{13-x}=\sqrt{x-5}\Leftrightarrow x=9\)

Vậy MaxA=4 khi x=9

Answer 2

\(A>0;A^2=\left(\sqrt{13-x}+\sqrt{x-5}\right)^2\le\left(1^2+1^2\right)\left(\sqrt{13-x}^2+\sqrt{x-5}^2\right)=2\left(13-x+x-5\right)=16.\)

0<A</ 4  =>  Max A = 4 khi 13-x = x -5 => x = 9