\(A=x^2+3x+7\)
\(A=x^2+2\cdot\dfrac{3}{2}\cdot x+\dfrac{9}{4}+\dfrac{19}{4}\)
\(A=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\)
Mà: \(\left(x+\dfrac{3}{2}\right)^2\ge0\)
\(\Rightarrow A=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)
Dấu "=" xảy ra khi:
\(x+\dfrac{3}{2}=0\Rightarrow x=-\dfrac{3}{2}\)
Vậy: \(A_{min}=\dfrac{19}{4}.khi.x=-\dfrac{3}{2}\)
\(B=\left(x-2\right)\left(x-4\right)+3\)
\(B=x^2-4x-2x+8+3\)
\(B=x^2-6x+11\)
\(B=\left(x-3\right)^2+2\)
Mà: \(\left(x-3\right)^2\ge0\)
\(\Rightarrow B=\left(x-3\right)^2+2\ge2\)
Dấu "=" xảy ra:
\(x-3=0\Rightarrow x=3\)
Vậy: \(B_{min}=2.khi.x=3\)
\(C=11-10x-x^2\)
\(C=-x^2-10x-25+36\)
\(C=-\left(x+5\right)^2+36\)
Mà: \(-\left(x+5\right)^2\le0\)
\(\Rightarrow C=-\left(x+5\right)^2+36\le36\)
Dấu "=" xảy ra:
\(x+5=0\Rightarrow x=-5\)
Vậy: \(C_{max}=36.khi.x=-5\)
