Đặt A=tử, B=mẫu
A=\(\left(x^3+3x+\frac{3}{x}+\frac{1}{x^3}\right)^2-\left(x^6+\frac{1}{x^6}\right)-2=x^6+\frac{1}{x^6}+9x^2+\frac{9}{x^2}+6x^4+\frac{6}{x^4}+18+2-x^6-\frac{1}{x^6}-2=6\left(x^4+\frac{1}{x^4}\right)+9\left(x^2+\frac{1}{x^2}\right)+18=6\left(\left(x^2+\frac{1}{x^2}\right)^2-2\right)+9\left(x^2+\frac{1}{x^2}\right)+18=6\left(\left(\left(x+\frac{1}{x}\right)^2-2\right)^2-2\right)+9\left(\left(x+\frac{1}{x}\right)^2-2\right)+18\)B=\(2\left(x^3+\frac{1}{x^3}\right)+3\left(x+\frac{1}{x}\right)=2\left(\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)\right)+3\left(x+\frac{1}{x}\right)\)
Đặt \(t=x+\frac{1}{x}\). Ta có
A=\(6\left(\left(t^2-2\right)^2-2\right)+9\left(t^2-2\right)+18\) \(=6\left(t^2-2\right)^2+9\left(t^2-2\right)-12+18=3\left(t^2-2\right)\left(2t^2-4+3\right)+6=\left(t^2-2\right)\left(2t^2-1\right)+6=2t^4-5t^2+2+6=2t^4-5t^2+7\)
Và B=\(2\left(t^3-3t\right)+3t=2t^3+3t\)
Vậy A/B=.....
