\(A=4x^2+4x+9\)

\(A=4\left(x^2+x+\dfrac{9}{4}\right)\)

\(A=4\left(x^2+2\cdot x\cdot0,5+0,25+2\right)\)

\(A=4\left(x+0,5\right)^2+8\)

Vì \(4\left(x+0,5\right)^2\ge0\forall x\)

\(\Rightarrow4\left(x+0,5\right)^2+8\ge8\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=-0,5\)

Vậy \(MIN_A=8\Leftrightarrow x=-0,5\)

\(A=4x^2+4x+9\)

\(A=4x^2+4x+1+8\)

\(A=\left(2x+1\right)^2+8\)

Vì (2x+1)² + 8 \(\ge\) 0 \(\forall x\) => minA= 8

Dấu "x" xảy ra <=> 2x + 1 = 0

2x = 0 -1

2x = -1

x = \(\dfrac{-1}{2}\)

Vậy minA = 8 khi x = \(-\dfrac{1}{2}\)

// cậu tham khảo //

a) \(6xy+4x-9y-7=0\)

  \(\Leftrightarrow2x.\left(3y+2\right)-9y-6-1=0\)

\(\Leftrightarrow2x.\left(3y+x\right)-3.\left(3y+2\right)=1\)

\(\Leftrightarrow\left(2x-3\right).\left(3y+2\right)=1\)

Mà \(x,y\in Z\Rightarrow2x-3;3y+2\in Z\)

Tự làm típ

\(A=x^3+y^3+xy\)

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

\(A=x^2-xy+y^2+xy\)( vì \(x+y=1\))

\(A=x^2+y^2\)

Áp dụng bất đẳng thức Bunhiakovxky ta có :

\(\left(1^2+1^2\right)\left(x^2+y^2\right)\ge\left(x\cdot1+y\cdot1\right)^2=\left(x+y\right)^2=1\)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge1\)

\(\Leftrightarrow x^2+y^2\ge\frac{1}{2}\)

Hay \(x^3+y^3+xy\ge\frac{1}{2}\)

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

A = 9x² + 6x + 15

A = [(3x + 6x + 1] + 14

A = (3x + 1)² + 14 \(\ge\)14

Dấu = xảy ra \(\Leftrightarrow\)3x + 1 = 0

                        \(\Rightarrow\)3x = - 1

                       \(\Rightarrow\)x = - 1 / 3

Min A = 14 \(\Leftrightarrow\)x = - 1 / 3

2, TC: \(\frac{5x^2-4x+4}{x^2}=\frac{4x^2+x^2-4x+4}{x^2}\)\(=\frac{4x^2}{x^2}+\frac{\left(x-2\right)^2}{x^2}=4+\frac{\left(x-2\right)^2}{x^2}\)

Ta có \(\frac{\left(x-2\right)^2}{x^2}\ge0\forall x\left(x\ne0\right)\)\(\Rightarrow4+\frac{\left(x-2\right)^2}{x^2}\ge4\)

Vậy GTNN của A là 4 tại \(\frac{\left(x-2^2\right)}{x^2}=0\Rightarrow x=2\)

Ta có: A = (x + 2)(x - 3)

= x² - x - 6

=\(x^2-2.\frac{1}{2}.x+\frac{1}{4}-\frac{25}{4}\)

= \(\left(x-\frac{1}{2}\right)^2-\frac{25}{4}\ge-\frac{25}{4}\)

Dấu "=" xảy ra <=> \(x-\frac{1}{2}=0\Rightarrow x=0,5\)

Vậy Min A = -25/4 <=> x = 0,5

a) Ta có :

\(A=2x-x^2-4\)

\(=2x-x^2-1-3\)

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

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

\(\Rightarrow Max_A=-3\Leftrightarrow x=1\)

Vậy ...

b) \(B=-x^2-4x\)

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

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

\(\Rightarrow Max_B=4\Leftrightarrow x=-2\)

Vậy ...