a: \(A=x^2+20x+100-91=\left(x+10\right)^2-91\ge-91\)
Dấu '=' xảy ra khi x=-10
b: \(B=3\left(x^2-\dfrac{4}{3}x+\dfrac{5}{3}\right)\)
\(=3\left(x^2-2\cdot x\cdot\dfrac{2}{3}+\dfrac{4}{9}+\dfrac{11}{9}\right)\)
\(=3\left(x-\dfrac{2}{3}\right)^2+\dfrac{11}{3}\ge\dfrac{11}{3}\)
Dấu '=' xảy ra khi x=2/3
\(A=x^2+20x+9=\left(x+10\right)^2-91\ge-91\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x+10\right)^2=0\Leftrightarrow x=-10\)
Vậy \(MinA=-91\)
\(B=3x^2-4x+5=3\left(x^2-\dfrac{4}{3}x\right)+5=3\left[\left(x-\dfrac{2}{3}\right)^2-\dfrac{4}{9}\right]+5=3\left(x-\dfrac{2}{3}\right)^2-\dfrac{4}{3}+5=\left(x-\dfrac{2}{3}\right)^2+\dfrac{11}{3}\ge\dfrac{11}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-\dfrac{2}{3}\right)^2=0\Leftrightarrow x=\dfrac{2}{3}\)
Vậy \(MinB=\dfrac{11}{3}\)
