Question

Tính giá trị biểu thức B=x⁹+\(\dfrac{1}{x^9}\) biết \(x+\dfrac{1}{x}=3\)

Answer 1

Lời giải:

\(B=x^9+\frac{1}{x^9}=\left(x^3+\frac{1}{x^3}\right)^3-3.x^3.\frac{1}{x^3}\left(x^3+\frac{1}{x^3}\right)\)

\(=\left(x^3+\frac{1}{x^3}\right)^3-3\left(x^3+\frac{1}{x^3}\right)\)

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

\(=\left[(x+\frac{1}{x})^3-3(x+\frac{1}{x})\right]^3-3\left[(x+\frac{1}{x})^3-3(x+\frac{1}{x})\right]\)

\(=(3^3-3.3)^3-3(3^3-3.3)=5778\)