`a)`\(A=3-\sqrt{x^2-2x}\)
Ta có: \(\sqrt{x^2-2x}\ge0\)
`=>`\(A\le3\)
Dấu "=" xảy ra khi \(x^2-2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
`b)`\(B=1-\sqrt{x^2-2x+2}\)
Ta có: \(\sqrt{x^2-2x+2}=\sqrt{\left(x-1\right)^2+1}\ge1\)
\(\Rightarrow B\le1-1=0\)
Dấu "=" xảy ra khi \(\left(x-1\right)^2=0\)
\(\Leftrightarrow x=1\)
`c)`\(C=1+\sqrt{4x-x^2-2}\)
\(C=1+\sqrt{-\left(x^2-4x+4\right)+2}\)
\(C=1+\sqrt{-\left(x-2\right)^2+2}\)
Ta có: \(\sqrt{-\left(x-2\right)^2+2}\le\sqrt{2}\)
\(\Rightarrow C\le1+\sqrt{2}\)
Dấu "=" xảy ra khi \(\left(x-2\right)^2=0\)
`<=>x=2`
