\(x^2+\frac{1}{x^2}=7\Leftrightarrow x^2+2.x.\frac{1}{x}+\frac{1}{x^2}=9\)
\(\Leftrightarrow\left(x+\frac{1}{x}\right)^2=9\Leftrightarrow x+\frac{1}{x}=3\)
\(P=x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3x.\frac{1}{x}\left(x+\frac{1}{x}\right)=3^3-3.3=18\)
\(Q=\left(x^3+\frac{1}{x^3}\right)\left(x^2+\frac{1}{x^2}\right)-\left(x+\frac{1}{x}\right)=7.18-3=...\)
Ta có : \(x^2+\frac{1}{x^2}=7\)
=> \(x^2+\frac{2.x^2.1}{x^2}+\frac{1}{x^2}=7+2=9\)
=> \(\left(x+\frac{1}{x}\right)^2=9\)
=> \(x+\frac{1}{x}=\pm3\)
Mà x > 0
=> \(x+\frac{1}{x}=3\)
Ta có : \(P=x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)\left(x^2+\frac{x.1}{x}+\frac{1}{x^2}\right)\)
=> \(P=\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}+1\right)=3\left(7+1\right)=3.8=24\)
Ta có : \(Q=x^5+\frac{1}{x^5}\)
=> \(Q=x^5+\frac{1}{x^5}+\left(x+\frac{1}{x}\right)-\left(x+\frac{1}{x}\right)\)
=> \(Q=x^5+\frac{1}{x^5}+x^2.\frac{1}{x^2}\left(x+\frac{1}{x}\right)-\left(x+\frac{1}{x}\right)\)
=> \(Q=x^5+\frac{1}{x^5}+\frac{x^3}{x^2}+\frac{x^2}{x^3}-\left(x+\frac{1}{x}\right)\)
=> \(Q=x^3\left(x^2+\frac{1}{x^2}\right)+\frac{1}{x^3}\left(x^2+\frac{1}{x^2}\right)-\left(x+\frac{1}{x}\right)\)
=> \(Q=24.7-3=165\)
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