a, \(A=9x^2-6x+5\)
\(=\left(9x^2-6x+1\right)+4\)
\(=\left(3x-1\right)^2+4\)
ta có:
\(\left(3x-1\right)^2\ge0\forall x\Rightarrow\left(3x-1\right)^2+4\ge4\forall x\)
Vậy Min A = 4
Để A = 4 thì \(3x-1=0\Rightarrow x=\dfrac{1}{3}\)
\(b,B=4x^2-5x\)
\(=\left(4x^2-5x+\dfrac{25}{16}\right)-\dfrac{25}{16}\)
\(=\left(2x-\dfrac{5}{4}\right)^2-\dfrac{25}{16}\)
TA có:
\(\left(2x-\dfrac{5}{4}\right)^2\ge\forall x\Rightarrow\left(2x-\dfrac{5}{4}\right)^2-\dfrac{25}{16}\ge-\dfrac{25}{16}\forall x\)Vậy Min B = \(-\dfrac{25}{16}\)
Để B = \(-\dfrac{25}{16}\) thì \(2x-\dfrac{5}{4}=0\Rightarrow2x=\dfrac{5}{4}\Rightarrow x=\dfrac{5}{8}\)
\(c,C=3x^2-6x\)
\(=3\left(x^2-2x+1\right)-3\)
\(=3\left(x-1\right)^2-3\)
Ta có:
\(3\left(x-1\right)^2\ge0\forall x\Rightarrow3\left(x-1\right)^2-3\ge-3\)
vậy Min C = -3
Để C = -3 thì x-1=0 => x = 1
\(d,D=5x^2-15x\)
\(=5\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{45}{4}\)
\(=5\left(x-\dfrac{3}{2}\right)^2-\dfrac{45}{4}\)
Ta có:
\(5\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\Rightarrow5\left(x-\dfrac{3}{2}\right)^2-\dfrac{45}{4}\ge-\dfrac{45}{4}\)Vậy Min D = \(-\dfrac{45}{4}\)
Để \(D=-\dfrac{45}{4}\) thì \(x-\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)
\(e,E=x^2+3x+4\)
\(=\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{7}{4}\)
\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)
Vậy Min E = \(\dfrac{7}{4}\) khi \(x+\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)
\(f,F=2x^2-4x+7\)
\(=2\left(x^2-2x+1\right)+5\)
\(=2\left(x-1\right)^2+5\ge5\forall x\)
Vậy Min F = 5 khi x - 1 =0 => x = 1
\(g,2x^2-3x=2\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{9}{8}\)
\(=2\left(x-\dfrac{3}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\forall x\)
Vậy Min G = \(\dfrac{-9}{8}\) khi \(x-\dfrac{3}{4}=0\Rightarrow x=\dfrac{3}{4}\)
\(h,H=3x^2-4x=3\left(x^2-\dfrac{4}{3}x+\dfrac{4}{9}\right)-\dfrac{4}{3}\)
\(=3\left(x-\dfrac{2}{3}\right)^2-\dfrac{4}{3}\ge-\dfrac{4}{3}\forall x\)
Vậy Min H = \(-\dfrac{4}{3}\) khi \(x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)
