Ta có: \(A=x^2-6x+10=x^2-2.3.x+9+1=\left(x-3\right)^2+1\)
\(\left(x-3\right)^2\ge0\)
\(\Rightarrow A=\left(x-3\right)^2+1\ge1\)
Dấu " = " khi \(x-3=0\Rightarrow x=3\)
Vậy \(MIN_A=1\) khi x = 3
Ta có:A=\(x^2-6x+10\)
\(\Leftrightarrow A=x^2-2.3.x+3^2+1\)
\(\Leftrightarrow A=\left(x-3\right)^2+1\)
A đạt GTNN khi \(\left(x-3\right)^2=0\Leftrightarrow x=3\)
A = (x2 - 6x + 9) + 1
A = (x - 3)2 + 1
Ta có: (x - 3)2 \(\ge\) 0 với mọi x
=> (x - 3)2 + 1 \(\ge\) 1 với mọi x
Dấu "=" xảy ra <=> x - 3 = 0 => x = 3
Vậy MIN A = 1 <=> x = 3
