A = (x^2-6x+9)+1
= (x-3)^2+1 >= 1
Dấu "=" xảy ra <=> x-3=0 <=> x=3
Vậy GTNN của A = 1 <=> x=3
Tk mk nha
\(A=x^2-6x+10\)
\(=\left(x^2-3x\right)-\left(3x-9\right)+1\)
\(=x\left(x-3\right)-3\left(x-3\right)+1\)
\(=\left(x-3\right)\left(x-3\right)+1\)
\(=\left(x-3\right)^2+1\) \(\ge1\) (vi (x-3)2 \(\ge0\))
Dấu "=" xảy ra \(\Leftrightarrow\)\(x-3=0\)
\(\Leftrightarrow\) \(x=3\)
Vậy \(Min\) \(A=1\) \(\Leftrightarrow\)\(x=3\)