\(\text{Tử }=\left(x^2-8x+16\right)+\left(x-2-2\sqrt{x-2}.\sqrt{2}+2\right)+2022\)
\(=\left(x-4\right)^2+\left(\sqrt{x-2}-\sqrt{2}\right)^2+2022\ge2022\)
Dấu "=" xảy ra khi \(x-4=0\text{ và }\sqrt{x-2}=\sqrt{2}\Leftrightarrow x=4\).
\(\text{Mẫu }=\sqrt{x-3}+\sqrt{5-x}\)
Ta có: \(\left(a-b\right)^2\ge0\Rightarrow a^2+b^2-2ab\ge0\Rightarrow2\left(a^2+b^2\right)\ge a^2+b^2+2ab=\left(a+b\right)^2\)
Dấu "=" xảy ra khi a = b.
\(\Rightarrow\left(\sqrt{x-3}+\sqrt{5-x}\right)^2\le2\left(x-3+5-x\right)=4\)
\(\Rightarrow\sqrt{x-3}+\sqrt{5-x}\le2\)
Dấu "=" xả ra khi \(\sqrt{x-3}=\sqrt{5-x}\Leftrightarrow x=4\)
\(\Rightarrow Q\ge\frac{2022}{2}=1011\)
Dấu "=" xảy ra khi x = 4.
Vậy GTNN của Q là 1011 khi x = 4.
