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

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

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

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

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

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

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

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

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

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

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 \(\dfrac{1}{3}\le x\le\dfrac{2}{3}\)

\(a,x^2+12x+39=x^2+12x+36+3=\left(x+6\right)^2+3\ge3\forall x\)

Dấu = xảy ra \(\Leftrightarrow x+6=0\) 

\(\Leftrightarrow x=-6\)

Vậy ...

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

Dấu = xảy ra \(\Leftrightarrow3x-2=0\)

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

Vậy ...

Trả lời:

a, \(x^2+12x+39=x^2+2.x.6+36+3=\left(x+6\right)^2+3\ge3\forall x\)

Dấu "=" xảy ra khi x + 6 = 0 <=> x = - 6

Vậy GTNN của biểu thức bằng 3 khi x = - 6

b, \(9x^2-12x=\left(3x\right)^2-2.3x.2+4-4=\left(3x-2\right)^2-4\ge-4\forall x\)

Dấu "=" xảy ra khi 3x - 2 = 0 <=> x = 2/3

Vậy GTNN của biểu thức bằng - 4 khi x = 2/3

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)

\(A=4x-x^2+3=-\left(x^2-4x-3\right)=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

Daaus = xayr ra khi: x = 2

b) \(B=4x^2-12x+15=4\left(x^2-3x+9\right)-21=4\left(x-3\right)^2-21\ge-21\)

Dấu = xảy ra khi x = 3

c) \(C=4x^2+2y^2-4xy-4y+1=\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3=\left(2x-y\right)^2+\left(y-2\right)^2-3\ge-3\)

Dấu = xảy ra khi

2x = y và y = 2

=> x = 1 và y = 2

a) A = \(-x^2+4x+3=-\left(x-2\right)^2+7\le7\)

Dấu "=" <=> x = 2

b) \(4x^2-12x+15=\left(2x-3\right)^2+6\ge6\)

Dấu "=" xảy ra <=> \(x=\dfrac{3}{2}\)

c) \(4x^2+2y^2-4xy-4y+1\)

= \(\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3\)

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

Dấu "=" <=> \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

\(A=5x^2+9y^2-12xy+24x-48y+81\)

\(A=4x^2+x^2+9y^2-12xy+32x-48y-8x+16+1+64\)

\(A=(4x^2+9y^2+64-12xy+32x-48y)+\left(x^2-8x+16\right)+1\)

\(A=[\left(2x\right)^2+\left(3y\right)^2+\left(8\right)^2-2.2x.3y-2.3y.8+2.2x.8]+\left(x^2-8x+16\right)+1\)

\(A=\left(2x-3y+8\right)^2\left(x-4\right)^2+1\)

\(Do\) \(\left(2x-3y+8\right)^2\ge0\) \(và\) \(\left(x-4\right)^2\ge0\)

\(\Rightarrow A_{min}=1\)

\(Y=\sqrt{\left(3x+2\right)^2+7}\ge\sqrt{0+7}=\sqrt{7}\)

\(Y_{Min}=\sqrt{7}\Leftrightarrow3x+2=0\Leftrightarrow x=-\frac{2}{3}\)