Ta có : \(\frac{a^4-3a^2+1}{a^4-a^2-2a-1}\) \(=\frac{\left(a^4-2a^2+1\right)-a^2}{\left(a^4-a^3-a^2\right)+\left(a^3-a^2-a\right)+\left(a^2-a-1\right)}\)

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

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

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

\(\frac{a^4-3a^2+1}{a^4-a^2-2a-1}\)

Theo đề bài ta có :

Tử số : \(a^4-2a^2+1-a^2\)

\(=\left(a^2-1\right)^2-a^2\)

\(=\left(a^2-1+a\right)\left(a^2-1-a\right)\)

Mẫu số : \(a^4-\left(a^2+2a+1\right)\)

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

\(=\left(a^2+a+1\right)\left(a^2-a-1\right)\)

Phân thức bằng \(\frac{a^2+a-1}{a^2+a+1}\)với điều kiện  \(a^2-a-1\ne0\)

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

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

\(=\frac{2x+x^2+6a+3ax+2x-x^2-6a+3ax-2a}{\left(2-x\right)\left(2+x\right)}+a\)

\(=\frac{6ax+4x-2a+a\left(4-x^2\right)}{\left(2-x\right)\left(2+x\right)}=\frac{6ax+4x-2a+4a-a\cdot x^2}{\left(2-x\right)\left(2+x\right)}=\frac{-a\cdot x^2+6\cdot ax+2a}{\left(2-x\right)\cdot\left(2+x\right)}\)

\(-a\cdot x^2+6a\cdot x+2a\)

\(=-a\cdot\frac{a^2}{\left(3a+2\right)^2}+6a\cdot\frac{a}{3a+2}+2a=\frac{-a^3}{\left(3a+2\right)^2}+\frac{6a^2\left(3a+2\right)+2a\left(3a+2\right)^2}{\left(3a+2\right)^2}\)

\(=\frac{-a^3+18a^3+12a^2+2a\left(9a^2+12a+4\right)}{\left(3a+2\right)^2}\)

\(=\frac{17a^3+12a^2+18a^3+24a^2+8a}{\left(3a+2\right)^2}=\frac{35a^3+36a^2+8a}{\left(3a+2\right)^2}\)

\(\left(2-x\right)\left(2+x\right)\)

\(=\left(2-\frac{a}{3a+2}\right)\left(2+\frac{a}{3a+2}\right)=\frac{6a+4-a}{\left(3a+2\right)}\cdot\frac{6a+4+a}{3a+2}\)

\(=\frac{\left(5a+4\right)\left(7a+4\right)}{\left(3a+2\right)^2}\)

\(A=\frac{-a\cdot x^2+6\cdot ax+2a}{\left(2-x\right)\cdot\left(2+x\right)}\)

\(=\frac{35a^3+36a^2+8a}{\left(3a+2\right)^2}:\frac{\left(5a+4\right)\left(7a+4\right)}{\left(3a+2\right)^2}=\frac{35a^3+36a^2+8a}{\left(5a+4\right)\left(7a+4\right)}\)