`@`\(A=x^2+x=x^2+2x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Dấu "=" `<=>x=-1/2`
`@`\(B=x^2-4x+3=x^2-4x+4-1=\left(x-2\right)^2-1\ge-1\)
Dấu "=" `<=>x=2`
`@`\(C=x^2+5x-1=x^2+2x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2-\left(\dfrac{5}{2}\right)^2-1=\left(x+\dfrac{5}{2}\right)^2-\dfrac{21}{4}\ge-\dfrac{21}{4}\)
Dấu "=" `<=>x=-5/2`
`@`\(D=x^2+8x-3=x^2+8x+16-19=\left(x+4\right)^2-19\ge-19\)
Dấu "=" `<=>x=-4`
`@`\(E=4x^2+6x+1=4x^2+6x+\left(\dfrac{3}{2}\right)^2-\left(\dfrac{3}{2}\right)^2+1=\left(2x+\dfrac{3}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\)
Dấu "=" `<=>x=-3/4`
`@`\(F=9x^2-16x+7=9x^2-16x+\left(\dfrac{8}{3}\right)^2-\left(\dfrac{8}{3}\right)^2+7=\left(3x+\dfrac{8}{3}\right)^2+\dfrac{127}{9}\ge\dfrac{127}{9}\)
Dấu "=" `<=>x=1`
\(A=x^2+x+\dfrac{1}{4}-\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Dấu "=" xảy ra khi \(\left(x+\dfrac{1}{2}\right)^2=0\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy \(MinA=-\dfrac{1}{4}\).
\(B=x^2-4x+4-1=\left(x-2\right)^2-1\ge-1\)
Dấu "=" xảy ra khi \(\left(x-2\right)^2=0\Leftrightarrow x=2\)
Vậy \(MinB=-1\).
\(C=x^2+5x+\dfrac{25}{4}-\dfrac{29}{4}=\left(x+\dfrac{5}{2}\right)^2-\dfrac{29}{4}\ge-\dfrac{29}{4}\)
Dấu "=" xảy ra khi \(\left(x+\dfrac{5}{2}\right)^2=0\Leftrightarrow x=-\dfrac{5}{2}\)
Vậy \(MinC=-\dfrac{29}{4}\)
\(D=x^2+8x+16-19=\left(x+4\right)^2-19\ge-19\)Dấu "=" xảy ra khi \(\left(x+4\right)^2=0\Leftrightarrow x=-4\)
Vậy \(MinD=-19\).
\(E=4x^2+6x+\dfrac{9}{4}-\dfrac{5}{4}=\left(2x+\dfrac{3}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\)Dấu "=" xảy ra khi \(\left(2x+\dfrac{3}{2}\right)^2=0\Leftrightarrow x=-\dfrac{3}{4}\)
Vậy \(MinE=-\dfrac{5}{4}\)
\(F=9x^2-16x+\dfrac{64}{9}-\dfrac{1}{9}=\left(3x-\dfrac{8}{3}\right)^2-\dfrac{1}{9}\ge-\dfrac{1}{9}\)Dấu "=" xảy ra khi \(\left(3x-\dfrac{8}{3}\right)^2=0\Leftrightarrow x=\dfrac{8}{9}\)
Vậy \(MinF=-\dfrac{1}{9}\).
