Question

Cho \(x+\dfrac{1}{x}=a.\) Tính \(x^5+\dfrac{1}{x^5}\) theo a

Answer 1

Có: \(x^5+\dfrac{1}{x^5}=(x^2+\dfrac{1}{x^2}).(x^3+\dfrac{1}{x^3})-(\dfrac{1}{x^2}.x^3+\dfrac{1}{x^3}.x^2)\)

\(=(x^2+\dfrac{1}{x^2}).(x^3+\dfrac{1}{x^3})-(x+\dfrac{1}{x})\)

\(=\left[\left(x+\dfrac{1}{x}\right)^2-2.x.\dfrac{1}{x}\right].\left[\left(x+\dfrac{1}{x}\right)^3-3.x.\dfrac{1}{x}.\left(x+\dfrac{1}{x}\right)\right]-a\)

\(=\left(a^2-2\right).\left(a^3-3a\right)-a\)

\(=a^5-5a^3+6a-a\)

\(=a^5-5a^3+5a\)

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