a) Đặt \(A=\left|x-3\right|+10\)
Ta có: \(\left|x-3\right|\ge0\)
\(\Rightarrow A=\left|x-3\right|+10\ge10\)
Vậy \(MIN_A=10\) khi x = 3
b) Đặt \(B=-7+\left(x-1\right)^2\)
Ta có: \(\left(x-1\right)^2\ge0\)
\(\Rightarrow B=-7+\left(x-1\right)^2\ge-7\)
Vậy \(MIN_B=-7\) khi x = 1
