a/ \(A=x^2-4x+15\)
\(=x^2-4x+4+11\)
\(=\left(x-2\right)^2+11\)
Nhận xét : \(\left(x-2\right)^2\ge0\)
\(\Leftrightarrow\left(x-2\right)^2+11\ge11\)
\(\Leftrightarrow A\ge11\)
Dấu "=" xảy ra khi \(\left(x-2\right)^2=0\Leftrightarrow x=2\)
Vậy \(A_{Min}=11\Leftrightarrow x=2\)
b/ \(B=9x^2-3x+17\)
\(=9x^2-3x+\dfrac{1}{4}+\dfrac{67}{4}\)
\(=\left(3x-\dfrac{1}{2}\right)^2+\dfrac{67}{4}\)
Nhận xét : \(\left(3x-\dfrac{1}{2}\right)^2\ge0\)
\(\Leftrightarrow\left(3x-\dfrac{1}{2}\right)^2+\dfrac{67}{4}\ge\dfrac{67}{4}\)
\(\Leftrightarrow B\ge\dfrac{67}{4}\)
Dấu "=" xảy ra khi : \(x=\dfrac{1}{6}\)
Vậy...
a)\(A=x^2-4x+15=\left(x-2\right)^2+11\)
Vì \(\left(x-2\right)^2\ge0\) nên muốn \(x^2-4x+15\) có được GTNN thì \((x-2)^2=0\)
\(\Rightarrow Min_A=0+11=11\)