\(A=\frac{1}{4}-\left(x-\sqrt{x}+\frac{1}{4}\right)=\frac{1}{4}-\left(\sqrt{x}-\frac{1}{2}\right)^2\le\frac{1}{4}\)
\(\Rightarrow A_{max}=\frac{1}{4}\) khi \(x=\frac{1}{4}\), \(A_{min}\) ko tồn tại
\(B=\sqrt{2-\left(9x^2+6x+1\right)}-5=\sqrt{2-\left(3x+1\right)^2}-5\)
Do \(0\le\sqrt{2-\left(3x+1\right)^2}\le\sqrt{2}\)
\(\Rightarrow B_{max}=\sqrt{2}-5\) khi \(x=-\frac{1}{3}\)
\(B_{min}=-5\) khi \(\left(3x+1\right)^2=2\Rightarrow x=\frac{-1\pm\sqrt{2}}{3}\)
\(C=\sqrt{x-2}+\sqrt{4-x}\ge\sqrt{x-2+4-x}=\sqrt{2}\)
\(\Rightarrow C_{min}=\sqrt{2}\) khi \(\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
\(C\le\sqrt{\left(1+1\right)\left(x-2+4-x\right)}=2\)
\(\Rightarrow C_{max}=2\) khi \(x-2=4-x\Leftrightarrow x=3\)