\(A=\sqrt{14-x}+\sqrt{x-10}\)
Đk:\(10\le x\le14\)
\(A^2=\left(\sqrt{14-x}+\sqrt{x-10}\right)^2\)
\(=\left(14-x\right)+\left(x-10\right)+2\sqrt{\left(14-x\right)\left(x-10\right)}\)
\(=4+2\sqrt{\left(14-x\right)\left(x-10\right)}\)
\(\le4+\left(14-x\right)+\left(x-10\right)\) (BĐT AM-GM)
\(=4+4=8\Rightarrow A^2\le8\Rightarrow A\le\sqrt{8}\)