\(A=x^2-2x+10\)
\(A=\left(x^2-2x+1\right)+9\)
\(A=\left(x-1\right)^2+9\)
Mà \(\left(x-1\right)^2\ge0\)
\(\Rightarrow A\ge9\)
Dấu "=" xảy ra khi :
\(x-1=0\Leftrightarrow x=1\)
Vậy Min A = 9 khi x = 1
\(B=x^2-5x-7\)
\(B=\left(x^2-5x+\frac{25}{4}\right)-\frac{53}{4}\)
\(B=\left(x-\frac{5}{2}\right)^2-\frac{53}{4}\)
Mà \(\left(x-\frac{5}{2}\right)^2\ge0\)
\(\Rightarrow B\ge-\frac{53}{4}\)
Dấu "=" xảy ra khi :
\(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)
Vậy \(B_{Min}=-\frac{53}{4}\Leftrightarrow x=\frac{5}{2}\)
\(C=3x^2+3x-5\)
\(3C=9x^2+9x-15\)
\(3C=\left(9x^2+9x+\frac{9}{4}\right)-\frac{69}{4}\)
\(3C=\left(3x+\frac{3}{2}\right)^2-\frac{69}{4}\)
Mà \(\left(3x+\frac{3}{2}\right)^2\ge0\)
\(\Rightarrow3C\ge-\frac{69}{4}\)
\(\Leftrightarrow C\ge-\frac{23}{4}\)
Dấu "=" xảy ra khi :
\(3x+\frac{3}{2}=0\Leftrightarrow x=-\frac{1}{2}\)
Vậy ...