\(A=x^2-6x-4=x^2-6x+9-13=\left(x-3\right)^2-13\ge-13\)
Vậy \(A_{min}=-13\Leftrightarrow x=3\)
\(B=x^2-x+1=x^2-2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy \(B_{min}=\frac{3}{4}\Leftrightarrow x=\frac{1}{2}\)
\(C=5x^2+x-3=5\left(x^2+\frac{1}{5}x-\frac{3}{5}\right)\)
\(=5\left(x^2+2.\frac{1}{10}x+\frac{1}{100}-\frac{61}{100}\right)\)
\(=5\left[\left(x+\frac{1}{10}\right)^2-\frac{61}{100}\right]=5\left(x+\frac{1}{10}\right)^2-\frac{61}{20}\ge\frac{-61}{20}\)
Vậy \(C_{min}=\frac{-61}{20}\Leftrightarrow x=\frac{-1}{10}\)
\(D=-x^2+3x-1=-\left(x^2-3x+1\right)\)
\(=-\left(x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{5}{4}\right)\)
\(=-\left(\left(x-\frac{3}{2}\right)^2-\frac{5}{4}\right)\)
\(=-\left(x-\frac{3}{2}\right)^2+\frac{5}{4}\le\frac{5}{4}\)
Vậy \(D_{max}=\frac{5}{4}\Leftrightarrow x=\frac{3}{2}\)
\(E=-3x^2+4x+2=-3\left(x^2+\frac{4}{3}x+\frac{2}{3}\right)\)
\(=-3\left(x^2+2.\frac{2}{3}x+\frac{4}{9}+\frac{2}{9}\right)\)
\(=-3\left(\left(x+\frac{2}{3}\right)^2+\frac{2}{9}\right)\)
\(=-3\left(x+\frac{2}{3}\right)^2-\frac{2}{3}\le-\frac{2}{3}\)
Vậy \(E_{max}=\frac{-2}{3}\Leftrightarrow x=\frac{-2}{3}\)
Sửa câu E
\(E=-3x^2+4x+2=-3\left(x^2-\frac{4}{3}x-\frac{2}{3}\right)\)
\(=-3\left(x^2-\frac{4}{3}x-\frac{2}{3}\right)\)
\(=-3\left(x^2-2.\frac{2}{3}x+\frac{4}{9}-\frac{10}{9}\right)\)
\(=-3\left(\left(x-\frac{2}{3}\right)^2-\frac{10}{9}\right)\)
\(=-3\left(x-\frac{2}{3}\right)^2+\frac{10}{3}\le\frac{10}{3}\)
Vậy ...
