Áp dụng bđt bunhia copski ta có
\(\left(\sqrt{x-5}+\sqrt{23-x}\right)^2=\left(\sqrt{x-5}.1+\sqrt{23-x}.1\right)^2\le\left[\left(\sqrt{x-5}\right)^2+\left(\sqrt{23-x}\right)^2\right]\left(1^2+1^2\right)=\left(x-5+23-x\right).2=36\Leftrightarrow A^2\le36\Leftrightarrow A\le6\)(vì A>0)
Dấu '=' xảy ra khi \(\dfrac{\sqrt{x-5}}{1}=\dfrac{\sqrt{23-x}}{1}\Leftrightarrow x-5=23-x\Leftrightarrow2x=28\Leftrightarrow x=14\)
Vậy giá trị lớn nhất của A là 6
\(A=\sqrt{x-5}+\sqrt{23-x}\)
\(A^2=x-5+23-x+2\sqrt{\left(x-5\right)\left(23-x\right)}=18+2\sqrt{\left(x-5\right)\left(23-x\right)}\)ÁP dụng BĐT AM - GM: \(2\sqrt{\left(x-5\right)\left(23-x\right)}\le x-5+23-x=18\)
\(\Rightarrow A^2\le36\)
\(\Rightarrow A\le6\)
"=" \(\Leftrightarrow x=14\)
