Ta có :
\(x+\frac{1}{x}=a\)
\(\Leftrightarrow\left(x+\frac{1}{x}\right)^3=a^3\)
\(\Leftrightarrow x^3+3x^2.\frac{1}{x}+3x.\frac{1}{x^2}+\frac{1}{x^3}=a^3\)
\(\Leftrightarrow\left(x^3+\frac{1}{x^3}\right)+3\left(x+\frac{1}{x}\right)=a^3\)
\(\Leftrightarrow x^3+\frac{1}{x^3}+3a=a^3\)
\(\Leftrightarrow x^3+\frac{1}{x^3}=a^3-3a\)
Lại có : \(x+\frac{1}{x}=a\)
\(\Leftrightarrow\left(x+\frac{1}{x}\right)^2=a^2\)
\(\Leftrightarrow x^2+2.x.\frac{1}{x}+\frac{1}{x^2}=a^2\)
\(\Leftrightarrow x^2+\frac{1}{x^2}=a^2-2\)
Ta có : \(\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=\left(a^3-3a\right)\left(a^2-2\right)\)
\(\Leftrightarrow x^5+\frac{1}{x}+x+\frac{1}{x^5}=a^5-2a^3-3a^3+6a\)
\(\Leftrightarrow x^5+\frac{1}{x^5}=a^5-5a^3+5a\)
Vậy.....
