Question

Cho f(x)=\(x^{2n}-x^{2n-1}+.....+x^2-x+1\) (x\(\in\)N)

g(x)=\(-x^{2n+1}+x^{2n}-x^{2n-1}+...+x^2-x+1\) (x\(\in\)N)

Tính giá trị của hiệu f(x)-g(x) tại x=\(\dfrac{1}{10}\)

Answer 1

Ta có:

\(f\left(x\right)-g\left(x\right)=\left(x^{2n}-x^{2n-1}+...+x^2-x+1\right)-\left(-x^{2n+1}+x^{2n}-x^{2n-1}+...+x^2-x+1\right)\)

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

\(\Rightarrow f\left(\dfrac{1}{10}\right)-g\left(\dfrac{1}{10}\right)=\left(\dfrac{1}{10}\right)^{2n+1}\)

Vậy \(f\left(\dfrac{1}{10}\right)-g\left(\dfrac{1}{10}\right)=\left(\dfrac{1}{10}\right)^{2n+1}\)