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Đặt \(\hept{\begin{cases}\left(x+\frac{1}{x}\right)^3=a\\x^3+\frac{1}{x^3}=b\end{cases}}\)
Ta có
\(A=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+2+\frac{1}{x^6}\right)}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^3+\frac{1}{x^3}\right)^2}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}\)
\(=\frac{a^2-b^2}{a+b}=a-b\)
\(=\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)\)
\(=x^3+3\left(x+\frac{1}{x}\right)+\frac{1}{x^3}-\left(x^3+\frac{1}{x^3}\right)=\frac{3x^2+3}{x}\)
B=\(\frac{3\left(2x^8+5x^6+6x^4+5x^2+2\right)}{x\left(x^2+1\right)\left(2x^4+x^2+2\right)}\)
a) \(P=\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ac}{\left(b-c\right)\left(b-a\right)}+\frac{ab}{\left(c-a\right)\left(c-b\right)}\)
Đặt \(x=\frac{b}{c-a},y=\frac{c}{a-b},z=\frac{a}{b-c}\) , suy ra : \(P=-xy-yz-xz\)
Lại có : \(\left(x-1\right)\left(y-1\right)\left(z-1\right)=\left(x+1\right)\left(y+1\right)\left(z+1\right)\)
\(\Rightarrow xy+yz+xz=-1\Rightarrow P=1\)
\(Q=\frac{\left[\left(x+\frac{1}{x}\right)^2\right]^3-\left(x^3+\frac{1}{x^3}\right)^2}{\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)}=\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)\)
\(=3x+\frac{3}{x}=3\left(x+\frac{1}{x}\right)\)
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=.....