1. Ta có: \(\left|x-1\right|\ge0\)
\(\Rightarrow\left|x-1\right|+15\ge15\)
\(\Rightarrow A\ge15\)
Dấu "=" xảy ra \(\Leftrightarrow\left|x-1\right|=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy, MinA = 15 \(\Leftrightarrow x=1\)
2. Ta có: \(\left(x-1\right)^2\ge0\)
\(\Rightarrow2+\left(x-1\right)^2\ge2\)
\(\Rightarrow B\ge2\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy MinB = 2 \(\Leftrightarrow x=1\)
3. Ta có: \(\left\{{}\begin{matrix}\left|x-1\right|\text{}\ge x-1\\\left|x-2\right|=\left|2-x\right|\ge2-x\end{matrix}\right.\)
\(\Rightarrow\left|x-1\right|+\left|2-x\right|\ge\left(x-1\right)+\left(2-x\right)\)
\(\Rightarrow\left|x-1\right|+\left|x-2\right|\ge1\)
\(\Rightarrow C\ge1\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\2-x\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x\le2\end{matrix}\right.\)
\(\Leftrightarrow1\le x\le2\)
Vậy MinC = 1 \(\Leftrightarrow1\le x\le2\)
