Vì |x + 1| ≥ 0 => |x + 1| + 2 ≥ 2

=> E ≤ 1/2

Dấu bằng xảy ra <=> |x + 1| = 0

                           <=> x + 1 = 0

                           <=> x = -1

Vậy Max E = 1/2 khi x = -1 

a) \(-ĐKXĐ:x\ne\pm2;1\)

Rút gọn : \(A=\left(\frac{1}{x+2}-\frac{2}{x-2}-\frac{x}{4-x^2}\right):\frac{6\left(x+2\right)}{\left(2-x\right)\left(x+1\right)}\)

\(=\left(\frac{1}{x+2}+\frac{-2}{x-2}+\frac{x}{x^2-4}\right).\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)

\(=\left[\frac{x-2}{\left(x-2\right)\left(x+2\right)}+\frac{\left(-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x}{\left(x-2\right)\left(x+2\right)}\right]\)\(.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)

\(=\left[\frac{x-2-2x-4+x}{\left(x-2\right)\left(x+2\right)}\right].\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)

\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)\(=\frac{x+1}{\left(x+2\right)^2}\)

b) \(A>0\Leftrightarrow\frac{x+1}{\left(x+2\right)^2}>0\Leftrightarrow\orbr{\begin{cases}x+1< 0;\left(x+2\right)^2< 0\left(voly\right)\\x+1>0;\left(x+2\right)^2>0\end{cases}}\)

\(\Leftrightarrow x>1;x>-2\Leftrightarrow x>1\)

Vậy với mọi x thỏa mãn x>1 thì A > 0

c) Ta có : \(x^2+3x+2=0\Leftrightarrow x^2+x+2x+2=0\)

\(\Leftrightarrow x\left(x+1\right)+2\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)

Vậy x = -1;-2

Câu a hình như sai đề mk sửa nha

a)\(A=\left(2x+\frac{1}{3}\right)^4-1\)

         Vì \(\left(2x+\frac{1}{3}\right)^4\ge0\)

      Suy ra:\(\left(2x+\frac{1}{3}\right)^4-1\ge-1\)

                   Dấu = xảy ra khi \(2x+\frac{1}{3}=0\)

                                               \(2x=-\frac{1}{3}\)

                                                \(x=-\frac{1}{6}\)

Vậy Min A=-1 khi \(x=-\frac{1}{6}\)

b)\(B=-\left(\frac{4}{9}x-\frac{2}{15}\right)^6+3\)

    \(B=3-\left(\frac{4}{9}x-\frac{2}{15}\right)^6\)

           Vì \(-\left(\frac{4}{9}x-\frac{2}{15}\right)^6\le0\)

                     Suy ra:\(3-\left(\frac{4}{9}x-\frac{2}{15}\right)^6\le3\)

Dấu = xảy ra khi \(\frac{4}{9}x-\frac{2}{15}=0\)

                            \(\frac{4}{9}x=\frac{2}{15}\)

                            \(x=\frac{3}{10}\)

     Vậy Max B=3 khi \(x=\frac{3}{10}\)