\(A=\frac{x^2+2x+3}{x^2+2}\)

\(A=\frac{x^2+2+2x+1}{x^2+2}\)

\(A=\frac{x^2+2}{x^2+2}+\frac{2x+1}{x^2+2}\)

\(A=1+\frac{x^2+2-x^2+2x-1}{x^2+2}\)

\(A=1+\frac{x^2+2}{x^2+2}-\frac{x^2-2x+1}{x^2+2}\)

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

\(A=2-\frac{\left(x-1\right)^2}{x^2+2}\le2\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

\(A=\frac{x^2+2x+3}{x^2+2}=\frac{2x^2+4x+6}{2\left(x^2+2\right)}=\frac{\left(x^2+4x+4\right)+\left(x^2+2\right)}{2\left(x^2+2\right)}=\frac{\left(x+2\right)^2}{2\left(x^2+2\right)}+\frac{1}{2}\ge\frac{1}{2}\forall x\)

Dấu "=" xảy ra khi: \(x+2=0\Leftrightarrow x=-2\)

Vậy GTNN của A là \(\frac{1}{2}\) khi x = -2

\(A=\frac{x^2+2x+3}{x^2+4x+4}-\frac{2}{3}+\frac{2}{3}\)

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

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

\(\hept{\begin{cases}\left(x-1\right)^2\ge0\\\left(x+2\right)^2\ge0\end{cases}\Rightarrow\frac{\left(x-1\right)^2}{\left(x+2\right)^2}\ge0}\)

Dấu '' ='' xảy ra khi và chỉ khi  x=1

=> Min A =2/3 khi x=1

a) x² - 2x + 5 = (x - 1)² + 4 >= 4

Min là 4 khi x = 1

\(P=\frac{2005x+2006\sqrt{1-x^2}+2007}{\sqrt{1-x^2}}\)

\(=\frac{2006\left(1+x\right)+\left(1-x\right)}{\sqrt{1-x^2}}+2006\)

\(\ge\frac{2\sqrt{2006\left(1+x\right)\left(1-x\right)}}{\sqrt{1-x^2}}+2006=2\sqrt{2006}+2006\)

Dấu = xảy ra khi:

\(2006\left(1+x\right)=1-x\)

\(\Leftrightarrow x=-\frac{2005}{2007}\)

\(A=\frac{1}{2017}-\frac{2}{2017x}+\frac{1}{x^2}=\left(\frac{1}{2017}-\frac{1}{x}\right)^2+\frac{1}{2017}-\frac{1}{2017^2}=\left(\frac{1}{2017}-\frac{1}{x}\right)^2+\frac{2016}{2017^2}\)

\(\Rightarrow A\ge\frac{2016}{2017^2}\)Dấu "=" xảy ra khi \(\left(\frac{1}{2017}-\frac{1}{x}\right)^2=0\Rightarrow x=2017\)

Vây ......