a)\(C\left(x\right)=9x^2-6x+14\)

\(\Rightarrow C\left(x\right)=\left(3x\right)^2-2.3x+1+13\)

\(\Rightarrow C\left(x\right)=\left(3x-1\right)^2+13\ge13\)

Dấu "=" xảy ra khi

\(3x-1=0\Leftrightarrow x=\frac{1}{3}\)

\(\Rightarrow MinC\left(x\right)=13\Leftrightarrow x=\frac{1}{3}\)

\(b,D\left(x\right)=3x^2+12x+15\)

\(\Rightarrow D\left(x\right)=3x^2+6x+6x+12+3\)

\(\Rightarrow D\left(x\right)=3x\left(x+2\right)+6\left(x+2\right)+3\)

\(\Rightarrow D\left(x\right)=3\left(x+2\right)^2+3\ge3.Với\forall x\in R\)

Dấu "=" xảy ra khi

\(x+2=0\Leftrightarrow x=-2\)

\(\Rightarrow MinD\left(x\right)=3\Leftrightarrow x=-2\)

a) C(x)= 9x^2 -6x + 14

       = (3x -1)^2 +13 >  13

vậy C_{min = 13 <=> x= 1/3}

b)D(x)= 3x^2 +12x +15

          = 3(x^2 +4x +5)

          = 3((x+2)^2 +1)

          = 3(x+2)^2 +3 > 3

vậy D_min=3 <=> x=-2

a, \(A=-x^2-2x+3=-\left(x^2+2x-3\right)=-\left(x^2+2x+1-4\right)\)

\(=-\left(x+1\right)^2+4\le4\)

Dấu ''='' xảy ra khi x = -1 

Vậy GTLN là 4 khi x = -1 

b, \(B=-4x^2+4x-3=-\left(4x^2-4x+3\right)=-\left(4x^2-4x+1+2\right)\)

\(=-\left(2x-1\right)^2-2\le-2\)

Dấu ''='' xảy ra khi x = 1/2 

Vậy GTLN B là -2 khi x = 1/2 

c, \(C=-x^2+6x-15=-\left(x^2-2x+15\right)=-\left(x^2-2x+1+14\right)\)

\(=-\left(x-1\right)^2-14\le-14\)

Vâỵ GTLN C là -14 khi x = 1

Bài 8 : 

b, \(B=x^2-6x+11=x^2-6x+9+2=\left(x-3\right)^2+2\ge2\)

Dấu ''='' xảy ra khi x = 3

Vậy GTNN B là 2 khi x = 3 

c, \(x^2-x+1=x^2-x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Dấu ''='' xảy ra khi x = 1/2 

Vậy ...

c, \(x^2-12x+2=x^2-12x+36-34=\left(x-6\right)^2-34\ge-34\)

Dấu ''='' xảy ra khi x = 6

Vậy ...

1) \(M=9x^2-6x+6=\left(9x^2-6x+1\right)+5=\left(3x-1\right)^2+5\ge5\)

\(minM=5\Leftrightarrow x=\dfrac{1}{3}\)

2) \(M=5-2x-x^2=-\left(x^2+2x+1\right)+6=-\left(x+1\right)^2+6\le6\)

\(maxM=6\Leftrightarrow x=-1\)

3) \(N=5+6x-9x^2=-\left(9x^2-6x+1\right)+6=-\left(3x-1\right)^2+6\le6\)

\(maxN=6\Leftrightarrow x=\dfrac{1}{3}\)

a) A = x² + 12x + 39

= ( x² + 12x + 36 ) + 3

= ( x + 6 )² + 3 ≥ 3 ∀ x

Đẳng thức xảy ra ⇔ x + 6 = 0 => x = -6

=> MinA = 3 ⇔ x = -6

B = 9x² - 12x 

= 9( x² - 4/3x + 4/9 ) - 4

= 9( x - 2/3 )² - 4 ≥ -4 ∀ x

Đẳng thức xảy ra ⇔ x - 2/3 = 0 => x = 2/3

=> MinB = -4 ⇔ x = 2/3

b) C = 4x - x² + 1

= -( x² - 4x + 4 ) + 5

= -( x - 2 )² + 5 ≤ 5 ∀ x

Đẳng thức xảy ra ⇔ x - 2 = 0 => x = 2

=> MaxC = 5 ⇔ x = 2

D = -4x² + 4x - 3

= -( 4x² - 4x + 1 ) - 2

= -( 2x - 1 )² - 2 ≤ -2 ∀ x

Đẳng thức xảy ra ⇔ 2x - 1 = 0 => x = 1/2

=> MaxD = -2 ⇔ x = 1/2

Ta có A = x² + 12x + 39 = (x² + 12x + 36) + 3 = (x + 6)² + 3 \(\ge\)3

Dấu "=" xảy ra <=> x + 6 = 0

=> x = -6

Vậy Min A = 3 <=> x = -6

Ta có B = 9x² - 12x = [(3x)² - 12x + 4] - 4 =(3x - 2)² - 4 \(\ge\)-4

Dấu "=" xảy ra <=> 3x - 2 =0

=> x = 2/3

Vậy Min B = -4 <=> x = 2/3

b) Ta có C = 4x - x² + 1 = -(x² - 4x - 1) = -(x² - 4x + 4) + 5 = -(x - 2)² + 5 \(\le\)5

Dấu "=" xảy ra <=> x - 2 = 0

=> x = 2

Vậy Max C = 5 <=> x = 2

Ta có D = -4x² + 4x - 3 = -(4x² - 4x + 1) - 2 = -(2x - 1)² - 2 \(\le\)-2

Dấu "=" xảy ra <=> 2x - 1 = 0

=> x = 0,5

Vậy Max D = -2 <=> x = 0,5

\(A=\sqrt{1-6x+9x^2}+\sqrt{9x^2-12x+4}\)

\(A=\sqrt{1^2-2\cdot3x\cdot1+\left(3x\right)^2}+\sqrt{\left(3x\right)^2-2\cdot2\cdot3x+2^2}\)

\(A=\sqrt{\left(1-3x\right)^2}+\sqrt{\left(3x-2\right)^2}\)

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

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

\(A=\left|-1\right|=1\)

Dấu "=" xảy ra \(\left(1-3x\right)\left(3x-2\right)\ge0\)

\(\Rightarrow\dfrac{1}{3}\le x\le\dfrac{2}{3}\)

Vậy: \(A_{min}=1\) khi \(\dfrac{1}{3}\le x\le\dfrac{2}{3}\)

Tìm GTNN

A = x² - 10x + 3 = ( x² - 10x + 25 ) - 22 = ( x - 5 )² - 22 ≥ -22 ∀ x

Dấu "=" xảy ra khi x = 5

=> MinA = -22 <=> x = 5

B = 3x² + 7x - 2 = 3( x² + 7/3x + 49/36 ) - 73/12 = 3( x + 7/6 )² - 73/12 ≥ -73/12 ∀ x

Dấu "=" xảy ra khi x = -7/6

=> MinB = -73/12 <=> x = -7/6

Tìm GTLN

A = -9x² + 12x - 5 = -9( x² - 4/3x + 4/9 ) - 1 = -9( x - 2/3 )² - 1 ≤ -1 ∀ x

Dấu "=" xảy ra khi x = 2/3

=> MaxA = -1 <=> x = 2/3

B = -2x² - 3x + 7 = -2( x² + 3/2x + 9/16 ) + 65/8 = -2( x + 3/4 )² + 65/8 ≤ 65/8 ∀ x

Dấu "=" xảy ra khi x = -3/4

=> MaxB = 65/8 <=> x = -3/4

\(A=\left(x+3\right)^2+2\ge2\\ A_{min}=2\Leftrightarrow x=-3\\ B=\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{29}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{29}{4}\ge-\dfrac{29}{4}\\ B_{min}=-\dfrac{29}{4}\Leftrightarrow x=-\dfrac{3}{2}\\ C=\left(9x^2-12x+4\right)+2017=\left(3x-2\right)^2+2017\ge2017\\ C_{min}=2017\Leftrightarrow x=\dfrac{2}{3}\)

\(A=\left(x^2-4x+4\right)+4=\left(x-2\right)^2+4\ge4\)

\(minA=4\Leftrightarrow x=2\)

\(B=\left(4x^2-12x+9\right)+2=\left(2x-3\right)^2+2\ge2\)

\(minB=2\Leftrightarrow x=\dfrac{3}{2}\)

\(C=3\left(x^2+2x+1\right)-8=3\left(x+1\right)^2-8\ge-8\)

\(minC=-8\Leftrightarrow x=-1\)

\(D=-\left(x^2-2x+1\right)-4=-\left(x-1\right)^2-4\le-4\)

\(maxD=-4\Leftrightarrow x=1\)

\(E=-\left(4x^2-6x+\dfrac{9}{4}\right)-\dfrac{11}{4}=-\left(2x-\dfrac{3}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\)

\(maxA=-\dfrac{11}{4}\Leftrightarrow x=\dfrac{3}{4}\)

\(F=-2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{55}{8}=-2\left(x-\dfrac{1}{4}\right)^2-\dfrac{55}{8}\le-\dfrac{55}{8}\)

\(maxF=-\dfrac{55}{8}\Leftrightarrow x=\dfrac{1}{4}\)

\(G=\left(x^2-4xy+4y^2\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-2y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(maxG=\dfrac{3}{4}\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-\dfrac{1}{2}\end{matrix}\right.\)

\(H=-\left(x^2-2x+1\right)-\left(y^2+4y+4\right)+16=-\left(x-1\right)^2-\left(y+2\right)^2+16\le16\)

\(maxH=16\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)