Question

Cho \(x=\dfrac{1-\sqrt{3}}{2};y=\dfrac{-1-\sqrt{3}}{2}\). Tính \(x^{11}+y^{11}\)

Answer 1

Lời giải:

Ta có \(x+y=-\sqrt{3}; xy=\frac{1}{2}\)

\(x^{11}+y^{11}=(x^5+y^5)(x^6+y^6)-x^5y^5(x+y)=(x^5+y^5)(x^6+y^6)+\frac{\sqrt{3}}{32}\)

Nhận thấy:

\(x^2+y^2=(x+y)^2-2xy=3-2.\frac{1}{2}=2\)

\(x^3+y^3=(x+y)^3-3xy(x+y)=-3\sqrt{3}+\frac{3\sqrt{3}}{2}=-1,5\sqrt{3}\)

\(x^5+y^5=(x^2+y^2)(x^3+y^3)-x^2y^2(x+y)\)

\(=-3\sqrt{3}+\frac{1}{4}\sqrt{3}=\frac{-11}{4}\sqrt{3}\)

\(x^6+y^6=(x^3+y^3)^2-2(xy)^3=(-1,5\sqrt{3})^2-2.\frac{1}{8}=\frac{13}{2}\)

Do đó: \(x^{11}+y^{11}=\frac{-11}{4}\sqrt{3}.\frac{13}{2}+\frac{\sqrt{3}}{32}=\frac{-571}{32}\sqrt{3}\)