Question

Cho a = \(\frac{\sqrt{6}+\sqrt{2}}{2}\)và b=\(\frac{\sqrt{6}-\sqrt{2}}{2}\)

Tính: 1/a^5+ 1/b^5

Answer 1

\(a+b=\sqrt{6}\)

\(a.b=1\Rightarrow b=\frac{1}{a}\Rightarrow\left\{{}\begin{matrix}\frac{1}{a^5}=b^5\\\frac{1}{b^5}=a^5\end{matrix}\right.\) \(\Rightarrow\frac{1}{a^5}+\frac{1}{b^5}=a^5+b^5\)

\(a^2+b^2=\left(a+b\right)^2-2ab=6-2=4\)

\(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=6\sqrt{6}-3\sqrt{6}=3\sqrt{6}\)

\(\left(a^2+b^2\right)\left(a^3+b^3\right)=a^5+b^5+\left(ab\right)^2\left(a+b\right)\)

\(\Leftrightarrow12\sqrt{6}=a^5+b^5+1.\sqrt{6}\)

\(\Rightarrow a^5+b^5=11\sqrt{6}\)