\(A=\dfrac{5x^2}{x^2}-\dfrac{x}{x^2}+\dfrac{1}{x^2}=\dfrac{1}{x^2}-\dfrac{1}{x}+5=\left(\dfrac{1}{x^2}-\dfrac{1}{x}+\dfrac{1}{4}\right)+\dfrac{19}{4}=\left(\dfrac{1}{x}-\dfrac{1}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)

\(A_{min}=\dfrac{19}{4}\) khi \(\dfrac{1}{x}=\dfrac{1}{2}\Rightarrow x=2\)

Bài 2: 

a) Ta có: \(\left|2x-5\right|\ge0\forall x\)

\(\Leftrightarrow-\left|2x-5\right|\le0\forall x\)

\(\Leftrightarrow-\left|2x-5\right|+3\le3\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{2}\)

Ta có A = 2x² + 12x + 1 

= \(2\left(x^2+6x+\frac{1}{2}\right)=2\left(x^2+6x+9-\frac{17}{2}\right)=2\left(x+3\right)^2-17\ge-17\)

=> Min A = -17

Dấu "=" xảy ra <=> x + 3 = 0 

<=> x = -3

Vậy Min A = -17 <=> x = -3

b) Ta có B = x² + 3x + 2 

= \(x^2+2.\frac{3}{2}x+\frac{9}{4}-\frac{1}{4}=\left(x+\frac{3}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

=> Min B = -1/4

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

Vậy Min B = -1/4 <=> x = -3/2 

(x^2+y^2-12y-12x+36)+(5y^2-10y+5)+4=(x-y-6)^2+5(y-1)^2+4>=4

GTNN A=4

khi y=1

x=7

a)\(A=4x^2+4x+11\)

\(=4x^2+4x+1+10\)

\(=\left(2x+1\right)^2+10\ge10\)

Dấu = khi \(x=\frac{-1}{2}\)

Vậy MinA=10 khi \(x=\frac{-1}{2}\)

b)\(B=3x^2-6x+1\)

\(=3x^2-6x+3-2\)

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

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

Dấu = khi \(x=1\)

Vậy MinB=-2 khi \(x=1\)

c)\(C=x^2-2x+y^2-4y+6\)

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

\(=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\)

Dấu = khi \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

Vậy MinC=1 khi \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

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