a ) \(A=3x^2+5x-2\)
\(=3\left(x^2+\dfrac{5}{3}x-\dfrac{2}{3}\right)\)
\(=3\left(x^2+2x.\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{49}{36}\right)\)
\(=3\left[\left(x+\dfrac{5}{6}\right)^2-\dfrac{49}{36}\right]\)
\(=3\left(x+\dfrac{5}{6}\right)^2-\dfrac{49}{12}\ge-\dfrac{49}{12}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x+\dfrac{5}{6}=0\Leftrightarrow x=-\dfrac{5}{6}\)
Vậy Min A là : \(-\dfrac{49}{12}\Leftrightarrow x=-\dfrac{5}{6}\)
b ) \(B=3x^2-4x+1\)
\(=3\left(x^2-\dfrac{4}{3}x+\dfrac{1}{3}\right)\)
\(=3\left(x^2-2x.\dfrac{2}{3}+\dfrac{4}{9}-\dfrac{1}{9}\right)\)
\(=3\left[\left(x-\dfrac{2}{3}\right)^2-\dfrac{1}{9}\right]\)
\(=3\left(x-\dfrac{2}{3}\right)^2-\dfrac{1}{3}\ge-\dfrac{1}{3}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x-\dfrac{2}{3}=0\Leftrightarrow x=\dfrac{2}{3}\)
Vậy Min B là : \(-\dfrac{1}{3}\Leftrightarrow x=\dfrac{2}{3}\)
c ) \(C=-x^2-3x-2\)
\(=-\left(x^2+3x+2\right)\)
\(=-\left(x^2+2x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{1}{4}\right)\)
\(=-\left[\left(x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}\right]\)
\(=-\left(x+\dfrac{3}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x+\dfrac{3}{2}=0\Leftrightarrow x=-\dfrac{3}{2}\)
Vậy Max C là : \(\dfrac{1}{4}\Leftrightarrow x=-\dfrac{3}{2}\)
d ) \(D=-4-3x^2+2x\)
\(=-3\left(x^2-\dfrac{2}{3}x+\dfrac{4}{3}\right)\)
\(=-3\left(x^2-2x.\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{11}{9}\right)\)
\(=-3\left[\left(x-\dfrac{1}{3}\right)^2+\dfrac{11}{9}\right]\)
\(=-3\left(x-\dfrac{1}{3}\right)^2-\dfrac{11}{3}\le-\dfrac{11}{3}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\)
Vậy Max D là : \(-\dfrac{11}{3}\Leftrightarrow x=\dfrac{1}{3}\)
:D
A=3x2+5x-2=2x2+4x+2+x2 -4=2(x+1)2 +x2-4 >=-4
Vậy A min=-4
B=3x2-4x+1=2x2-4x+2+x2-1=2(x-1)2+x2-1>=-1
Vậy B min=-1
C=-x2-3x -2=-(x2+3x+2)=-(x2+2x.3/2+9/4-1/4)=-(x+3/2)2+1/4
Ta có -(x+3/2)2<=0
=>-(x+3/2)2+1/4<=1/4
=> C max=1/4
D=-4-3x2+2x=-3x2+2x-4=-3(x2-2x/3+4/3)
=-3(x2-2x.1/3+1/9+11/9)=-3(x-1/3)2-11/3
Ta có -3(x-1/3)2<=0
=>-3(x-1/3)2-11/3<=-11/3
Vậy D max=-11/3
\(a,A=3x^2+5x-2\)
\(=3\left(x^2+\dfrac{5}{3}x-\dfrac{2}{3}\right)\)
\(=3\left(x^2+2.x.\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{97}{36}\right)\)
\(=3\left[\left(x+\dfrac{5}{6}\right)^2-\dfrac{97}{36}\right]\)
\(=3\left(x+\dfrac{5}{6}\right)^2-\dfrac{97}{12}\ge-\dfrac{97}{12}\) (dấu "=" xảy ra \(\Leftrightarrow x=-\dfrac{5}{6}\))
Vậy \(A_{min}=-\dfrac{97}{12}\) tại \(x=-\dfrac{5}{6}\)
\(b,B=3x^2-4x+1\)
\(=3\left(x^2-\dfrac{4}{3}x+\dfrac{1}{3}\right)\)
\(=3\left(x^2-2.x.\dfrac{2}{3}+\dfrac{4}{9}-\dfrac{5}{9}\right)\)
\(=3\left[\left(x-\dfrac{2}{3}\right)^2-\dfrac{5}{9}\right]\)
\(=3\left(x-\dfrac{2}{3}\right)^2-\dfrac{5}{3}\ge-\dfrac{5}{3}\) (dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{2}{3}\))
Vậy \(B_{min}=-\dfrac{5}{3}\) tại \(x=\dfrac{2}{3}\)
\(c,C=-x^2-3x-2\)
\(=-\left(x^2+3x+2\right)\)
\(=-\left(x^2+2.x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{17}{4}\right)\)
\(=-\left[\left(x+\dfrac{3}{2}\right)^2-\dfrac{17}{4}\right]\)
\(=-\left(x+\dfrac{3}{2}\right)^2+\dfrac{17}{4}\le\dfrac{17}{4}\) (dấu "=" xảy ra \(\Leftrightarrow x=-\dfrac{3}{2}\))
Vậy \(C_{max}=\dfrac{17}{4}\) tại \(x=-\dfrac{3}{2}\)
\(d,D=-4-3x^2+2x\)
\(=-3x^2+2x-4\)
\(=-\left(3x^2-2x+4\right)\)
\(=-\left[3\left(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{15}{4}\right)\right]\)
\(=-\left\{3\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{15}{4}\right]\right\}\)
\(=-3\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{15}{4}\right]\)
\(=-3\left(x-\dfrac{1}{2}\right)^2+\dfrac{45}{4}\le\dfrac{45}{4}\) (dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\))
Vậy \(D_{max}=\dfrac{45}{4}\) tại \(x=\dfrac{1}{2}\)
