a: =-x^2+6x-4

=-(x^2-6x+4)

=-(x^2-6x+9-5)

=-(x-3)^2+5<=5

Dấu = xảy ra khi x=3

b: =3(x^2-5/3x+7/3)

=3(x^2-2*x*5/6+25/36+59/36)

=3(x-5/6)^2+59/12>=59/12

Dấu = xảy ra khi x=5/6

c: \(=-\left(x-3\right)^2+2\left|x-3\right|\)

\(=-\left[\left(\left|x-3\right|\right)^2-2\left|x-3\right|+1-1\right]\)

\(=-\left(\left|x-3\right|-1\right)^2+1< =1\)

Dấu = xảy ra khi x=4 hoặc x=2

a: \(f\left(x\right)=2x^2-7x+9\)

=>\(f'\left(x\right)=2\cdot2x-7=4x-7\)

Đặt f'(x)=0

=>\(4x-7=0\)

=>\(x=\dfrac{7}{4}\)

\(f\left(\dfrac{7}{4}\right)=2\cdot\left(\dfrac{7}{4}\right)^2-7\cdot\dfrac{7}{4}+9=\dfrac{23}{8}\)

\(f\left(-1\right)=2\left(-1\right)^2-7\cdot\left(-1\right)+9=18\)

\(f\left(4\right)=2\cdot4^2-7\cdot4+9=13\)

Vì \(f\left(\dfrac{7}{4}\right)< f\left(4\right)< f\left(-1\right)\)

nên \(f\left(x\right)_{max\left[-1;4\right]}=18;f\left(x\right)_{min\left[-1;4\right]}=\dfrac{23}{8}\)

b: \(f\left(x\right)=x^2+5x+3\)

=>\(f'\left(x\right)=2x+5\)

f'(x)=0

=>2x+5=0

=>2x=-5

=>\(x=-\dfrac{5}{2}\)

\(f\left(-\dfrac{5}{2}\right)=\left(-\dfrac{5}{2}\right)^2+5\cdot\dfrac{-5}{2}+3=\dfrac{25}{4}-\dfrac{25}{2}+3=-\dfrac{13}{4}\)

\(f\left(2\right)=2^2+5\cdot2+3=4+10+3=17\)

\(f\left(6\right)=6^2+5\cdot6+3=69\)

Vậy: \(f\left(x\right)_{max\left[2;6\right]}=69;f\left(x\right)_{min\left[2;6\right]}=-\dfrac{13}{4}\)

A = -4 - x² + 6x = -(x² - 6x + 9) + 5 = -(x - 3)² + 5 \(\le\)5 \(\forall\) x

Dấu "=" xảy ra <=> x - 3  = 0 <=> x = 3

Vậy MaxA = 5 khi x = 3

F = (x - 1)(x - 3) + 11 = x² - 4x + 3 + 11 = (x² - 4x + 4) + 10 = (x - 2)² + 10 \(\ge\)10 \(\forall\)x

Dấu "=" xảy ra <=> x  - 2 = 0 <=> x = 2

Vậy MinF = 10 khi x = 2

B = 3x² - 5x + 7 = 3(x² - 5/3x + 25/36) + 59/12 = 3(x - 5/3)² + 59/12 \(\ge\)59/12 \(\forall\)x

Dấu "=" xảy ra <=> x - 5/3 = 0 <=>  x = 5/3

Vậy MinB = 59/12 khi x = 5/3

G = (x - 3)² + (x - 2)² = x² - 6x + 9 + x² - 4x + 4 = 2x² - 10x + 13 = 2(x² - 5x + 25/4) + 1/2 = 2(x - 5/2)² + 1/2 \(\ge\)1/2 \(\forall\)x

Dấu "=" xảy ra <=> x - 5/2 = 0 <=> x = 5/2

Vậy MinG = 1/2 khi x  = 5/2