Question

Cho \(x+\frac{1}{x}=4\) (với x ≠ 0)

a) Tính \(x^2+\frac{1}{x^2}\)

b) Tính \(x^3+\frac{1}{x^3}\)

c) Tính \(x^5+\frac{1}{x^5}\)

d) Tính \(x^6+\frac{1}{x^6}\)

Answer 1

\(a,x+\frac{1}{x}=4\Rightarrow\left(x+\frac{1}{x}\right)^2=x^2+2+\frac{1}{x^2}=16\Rightarrow x^2+\frac{1}{x^2}=14\)

\(b,\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}\right)=x^3+\frac{1}{x}+x+\frac{1}{x^3}=\left(x^3+\frac{1}{x^3}\right)+\left(x+\frac{1}{x}\right)=\left(x^3+\frac{1}{x^3}\right)+4=4.14=56\Rightarrow x^3+\frac{1}{x^3}=52\) \(c,\left(x^3+\frac{1}{x^3}\right)\left(x^2+\frac{1}{x^2}\right)=x^5+x+\frac{1}{x}+\frac{1}{x^5}=\left(x^5+\frac{1}{x^5}\right)+\left(x+\frac{1}{x}\right)=x^5+\frac{1}{x^5}+4=52.14=728\Rightarrow x^5+\frac{1}{x^5}=724\) \(d,\left(x^3+\frac{1}{x^3}\right)^2=52^2=x^6+2+\frac{1}{x^6}=x^6+\frac{1}{x^6}+2=2704\Rightarrow x^6+\frac{1}{x^6}=2702\)