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 \(\dfrac{1}{3}\le x\le\dfrac{2}{3}\)

A=9x^2+18xy-12x+13y^2-24y+5

\(=\left(3x\right)^2+2.3.3xy-2.3x.2+9y^2+4y^2-12y-12y+4+9-8\)

\(=\left[\left(3x\right)^2+\left(3y\right)^2+2^2+2.3x.3y+2.3x.2+2.3y.2\right]+\left[\left(2y\right)^2-2.2y.3+9\right]-8\)

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

Vậy \(MinA=-8\Leftrightarrow\hept{\begin{cases}\left(3x+3y+2\right)^2=0\\\left(2y-3\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}3x+3y+2=0\\2y-3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-6,5\\y=1,5\end{cases}}}\)

a) \(A=x^2-2x+6\\ =\left(x^2-2x+1\right)+5\\ =\left(x-1\right)^2+5\ge5\)

Dấu "=" xảy ra \(\Leftrightarrow x=1\)

b) \(B=4x^2+12x-3\\ =\left(4x^2+12x+9\right)-6\\ =\left(2x+3\right)^2-6\ge-6\)

Dấu "=" xảy ra \(\Leftrightarrow x=-\dfrac{3}{2}\)

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

Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\)

\(A=x^2-2x+6=\left(x-1\right)^2+5\ge5\)

\(minA=5\Leftrightarrow x=1\)

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

\(minB=-12\Leftrightarrow x=-\dfrac{3}{2}\)

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

\(minC=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}\)

a) \(x^2\)\(+3x+7\)

=\(x^2\)\(+2.x.\frac{3}{2}\)\(+\frac{9}{4}\)\(+\frac{19}{4}\)

=\(\left(x+\frac{3}{2}\right)^2\)\(+\frac{19}{4}\)

Vì \(\left(x+\frac{3}{2}\right)^2\)\(\ge0\)

Nên \(\left(x+\frac{3}{2}\right)^2\)\(+\frac{19}{4}\)\(\ge\frac{19}{4}\)

Dấu "=" xảy ra khi:

 \(x+\frac{3}{2}\)\(=0\)

\(\Rightarrow x=-\frac{3}{2}\)

Vậy GTNN của \(x^2\)\(+3x+7\) là \(\frac{19}{4}\) khi \(x=-\frac{3}{2}\)

b) \(-9x^2+12x-15\)

=\(-\left(9x^2-12x+15\right)\)

=\(-\left(\left(3x\right)^2-2.3x.2+4+11\right)\)

=\(-\left(\left(3x-2\right)^2+11\right)\)

=\(-\left(3x-2\right)^2-11\)

Vì \(\left(3x-2\right)^2\)\(\ge0\)

Nên \(-\left(3x-2\right)^2-11\le-11\)

Dấu "=" xảy ra khi:

\(3x-2=0\)

\(\Rightarrow x=\frac{2}{3}\)

Vậy GTLN của \(-9x^2+12x-15\) là \(-11\) khì \(x=\frac{2}{3}\)

c) \(11-10x-x^2\)

=\(-\left(x^2+10x-11\right)\)

=\(-\left(x^2+2.x.5+25-36\right)\)

=\(-\left(\left(x+5\right)^2-36\right)\)

=\(-\left(x+5\right)^2+36\)

Vì \(\left(x+5\right)^2\ge0\)

Nên \(-\left(x+5\right)^2+36\le36\)

Dấu "=" xảy ra khi:

 \(x+5=0\)

\(\Rightarrow x=-5\)

Vậy GTLN \(11-10x-x^2\) là \(36\) khi \(x=-5\)

d)\(x^4+x^2+2\)

=\(\left(x^2\right)^2+2.x^2.\frac{1}{2}+\frac{1}{4}+\frac{7}{4}\)

=\(\left(x^2+\frac{1}{2}\right)^2+\frac{7}{4}\)

Vì \(\left(x^2+\frac{1}{2}\right)^2\ge0\)

Nên \(\left(x^2+\frac{1}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}\)

Dấu "=" xảy ra khi:

 \(x^2+\frac{1}{2}=0\)

\(\Rightarrow x=\frac{1}{\sqrt{2}}\)

Vậy GTNN của \(x^4+x^2+2\) là \(\frac{7}{4}\) khi \(x=\frac{1}{\sqrt{2}}\)

a) \(x^2+3x+7=x^2+2.1,5x+1,5^2+4,75=\left(x+1,5\right)^2+4,75\ge4,75\)

Đẳng thức xảy ra khi : \(x+1,5=0\Rightarrow x=-1,5\)

Vậy giá trị nhỏ nhất của x² + 3x + 7 là 4,75 khi x = -1,5

b) \(-9x^2+12x-15=-\left(9x^2-12x+15\right)=-\left[\left(3x\right)^2-2.2.3x+2^2+11\right]\)

\(=-\left[\left(3x-2\right)^2+11\right]=-\left(3x-2\right)^2-11\le-11\)

Đẳng thức xảy ra khi :  \(3x-2=0\Rightarrow x=\frac{2}{3}\)

Vậy giá trị lớn nhất của -9x² +12x - 15 là -11 khi \(x=\frac{2}{3}\)

c) \(11-10x-x^2=-x^2-10x+11=-\left(x^2+10x-11\right)=-\left(x^2+2.5x+5^2-36\right)\)

\(=-\left[\left(x+5\right)^2-36\right]=-\left(x+5\right)^2+36\le36\)

Đẳng thức xảy ra khi : \(x+5=0\Rightarrow x=-5\)

Vậy giá trị lớn nhất của 11 - 10x -x² là 36 khi x = -5. 

\(\text{A=9x^2+18xy-12x+13y^2-24y+5}\)

\(=\left[\left(3x\right)^2+\left(3y\right)^2+2^2-12x+18xy-12y\right]+\left[\left(2y\right)^2-2.2y.3+9\right]-8\)

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

Vậy \(MinA=-8\Leftrightarrow\hept{\begin{cases}3x+3y-2=0\\2y-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1,5\\y=1,5\end{cases}}}\)