\(\sqrt{x-3}+\sqrt{13-x}=2\sqrt{5}\)
Áp dụng BĐT Bu-nhi-a ta có:
\(\left(\sqrt{x-3}+\sqrt{13-x}\right)^2\le\left(1^2+1^2\right)\left(\sqrt{x-3}^2+\sqrt{13-x}^2\right)=2\sqrt{5}\)Dấu ''='' xảy ra khi
\(\sqrt{x-3}=\sqrt{13-x}\)
\(\Leftrightarrow x-3=13-x\)
\(\Leftrightarrow x=8\)
Vậy...........