Áp dụng bất đẳng thức Bunyakovsky ta có:
\(A=\sqrt{13-x}+\sqrt{x-5}\)
\(A^2=\left(\sqrt{13-x}+\sqrt{x-5}\right)^2\)
\(\le\left(1^2+1^2\right)\left(13-x+x-5\right)=16\)
\(\Rightarrow A^2\le16\Leftrightarrow A\le4\)
Dấu "=" xảy ra khi:
\(\dfrac{1}{\sqrt{13-x}}=\dfrac{1}{\sqrt{x-5}}\Leftrightarrow\sqrt{13-x}=\sqrt{x-5}\Leftrightarrow13-x=x-5\)
\(\Rightarrow13=2x-5\Leftrightarrow2x=18\Leftrightarrow x=9\)
\(B=\sqrt{x-5}+\sqrt{37-x}\)
\(B^2=\left(\sqrt{x-5}+\sqrt{37-x}\right)^2\)
\(\le\left(1^2+1^2\right)\left(x-5+37-x\right)=64\)
\(\Rightarrow B^2\le64\Leftrightarrow B\le8\)
Dấu "=" xảy ra khi:
\(\dfrac{1}{\sqrt{x-5}}=\dfrac{1}{\sqrt{37-x}}\Leftrightarrow\sqrt{x-5}=\sqrt{37-x}\)
\(\Rightarrow x-5=37-x\Leftrightarrow2x-5=37\Leftrightarrow2x=42\Leftrightarrow x=21\)
