\(E=\frac{x^2}{x-2}.\left(\frac{x^2+4}{x}-4\right)+3\)( \(ĐK:x\ne2;x\ne0\))

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

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

b, \(E=x^2-2x+3=\left(x-1\right)^2+2\ge2\forall x\)

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

Vậy GTNN của E là 2 khi x = 1

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

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

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

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

Dấu '' ='' xảy ra khi và chỉ khi  x=1

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

Có \(A=\frac{2x+1}{x^2+3}\)

\(\Leftrightarrow Ax^2+3A=2x+1\)

\(\Leftrightarrow Ax^2-2x+3A-1=0\)

Có \(\Delta'=1-A\left(3A-1\right)\)

         \(=1-3A^2+A\)

Pt có nghiệm khi \(\Delta'\ge0\Leftrightarrow-3A^2+A+1\ge0\)

                                        \(\Leftrightarrow\frac{1-\sqrt{13}}{6}\le A\le\frac{1+\sqrt{13}}{6}\)

Nên \(A_{min}=\frac{1-\sqrt{13}}{6}\)

Dấu "=" \(\Leftrightarrow\frac{2x+1}{x^2+3}=\frac{1-\sqrt{13}}{6}\)

Giải ra tìm đc x

Vậy .............

\(A=\frac{16x}{3-x}+\frac{3}{x}+1=\frac{16x}{3-x}+\frac{3-x}{x}+2\ge8+2=10\)

Dau '=' xay ra khi \(x=\frac{3}{5}\)

Vay \(A_{min}=10\)khi \(x=\frac{3}{5}\)

Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) , có :

\(A=\left|x-3\right|+\left|x+\frac{3}{2}\right|\)

\(\Leftrightarrow A=\left|x-3\right|+\left|-x-\frac{3}{2}\right|\ge\left|x-3-x-\frac{3}{2}\right|=\left|-\frac{9}{2}\right|=\frac{9}{2}\)

\(\Rightarrow Min=\frac{9}{2}\)

\(\Leftrightarrow\hept{\begin{cases}x-3\le0\\x+\frac{3}{2}\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le3\\x\ge-\frac{3}{2}\end{cases}}\Rightarrow\frac{-3}{2}\le x\le3\)

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

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

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

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

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

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

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

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