Đáp án:
\(B_{\min}=2\Leftrightarrow x=4\)
Giải thích các bước giải:
\(\begin{split}B=\sqrt{x-3}+\sqrt{5-x}\ \left(3\le x\le 5\right)\\\Rightarrow B^2=\left(\sqrt{x-3}+\sqrt{5-x}\right)^2&\overset{\text{bunhia}}{\le} 2\left(\left(\sqrt{x-3}\right)^2+\left(\sqrt{5-x}\right)^2\right)\end{split}\\ \Rightarrow B^2\le 2\left(x-3+5-x\right)=4\Leftrightarrow B\le 2\\\Rightarrow B_{\min}=2\Leftrightarrow x-3=5-x\\\Leftrightarrow x=4\ (TM)\)