\(A=x^2-12x+18\)
\(A=x^2-2.x.6+36-36+18\)
\(A=\left(x-6\right)^2-18\)
Vì \(\left(x-6\right)^2\ge0\)
Nên \(\left(x-6\right)^2-18\ge-18\)
Vậy \(A_{MIN}=-18\Leftrightarrow x-6=0\Leftrightarrow x=6\)
Ta có : \(A=x^2-12x+18\)
\(=x^2-2.x.6+6^2-18\)
\(=\left(x-6\right)^2-18\)
Có : \(\left(x-6\right)^2\ge0\)
\(\Rightarrow\left(x-6\right)^2-18\ge-18\)
Dấu " = " xảy ra khi \(x-6=0\)
\(x=6\)
Vậy \(MIN_A=-18\) khi \(x=6\)
A= x2 - 12x + 18
=x2 - 12x + 36-18
=(x-6)2-18\(\ge\)-18
Dấu = khi x=6
Vậy MinA=-18 khi x=6
\(A=x^2-12x+18=x^2-2\times x\times6+36-36+18=\left(x-6\right)^2-18\)
\(\left(x-6\right)^2\ge0\)
\(\left(x-6\right)^2-18\ge-18\)
Vậy Min A = - 18 khi x = 6
A = x2-2.6x + 36 -18
A = (x-6)2 -18
vì (x-6)2 >= 0
=> (x-6)2 - 18 >= -18
Vậy Min A = -18 (=) x-6 = 0
=> x=6
