\(a,\)

\(A=\left(\frac{4x}{x+2}-\frac{x^3-8}{x^3+8}.\frac{4x^2-4x+16}{x^2-4}\right):\frac{16}{x+2}.\frac{x^2+3x+2}{x^2+x+1}\)\(ĐKXĐ:x\ne\pm2\)

\(A=[\frac{4x}{x+2}-\frac{\left(x-2\right)\left(x^2+2x+4\right).4\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)\left(x-2\right)\left(x+2\right)}]:\frac{16}{x+2}.\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}\)

\(A=[\frac{4x}{x+2}-\frac{4\left(x^2+2x+4\right)}{\left(x+2\right)^2}].\frac{x+2}{16}.\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}\)

\(A=\frac{4x^2+8x-4x^2-8x-16}{\left(x+2\right)^2}.\frac{x+2}{16}.\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}\)

\(A=\frac{16\left(x+2\right)}{\left(x+2\right)^2.16}.\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}\)

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

\(B=\frac{x^2+x-2}{x^3-1}\)\(ĐKXĐ:x\ne1\)

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

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

\(b,\)

Ta có:

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

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

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

\(\Rightarrow A+B=\frac{1}{x^2+x+1}=\frac{1}{x^2+2.x.\left(\frac{1}{2}\right)^2+\frac{3}{4}}=\frac{1}{\left(x+\frac{1}{2}\right)^2}+\frac{3}{4}\)

Vì:\(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)

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

\(\Rightarrow\frac{1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}\le\frac{1}{\frac{3}{4}}\)

\(\Rightarrow A+B\le\frac{4}{3}\)

\(\Rightarrow GTLN\)của \(A+B=\frac{4}{3}\Leftrightarrow x+\frac{1}{2}=0\)

                                                        \(\Leftrightarrow x=\frac{-1}{2}\left(TMĐK\right)\)

Vậy........

\(P=\frac{4x-x^3-x+4x^3}{1-4x^2}:\frac{4x^2-x^4+1-4x^2}{1-4x^2}\)

\(=\frac{3x^3+3x}{1-4x^2}:\frac{1-x^4}{1-4x^2}\)

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

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

\(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)

\(=\left[\frac{x^2}{x\left(x^2-4\right)}+\frac{-6}{3\left(x-2\right)}+\frac{1}{x+2}\right]:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)

\(=\left[\frac{x}{\left(x-2\right)\left(x+2\right)}+\frac{-2}{x-2}+\frac{1}{x+2}\right]:\left[\frac{x^2-4}{x+2}+\frac{10-x^2}{x+2}\right]\)

\(=\left[\frac{x}{\left(x-2\right)\left(x+2\right)}+\frac{-2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)

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