Có: \(a+b=\dfrac{\sqrt{6}+\sqrt{2}}{2}+\dfrac{\sqrt{6}-\sqrt{2}}{2}=\dfrac{2\sqrt{6}}{2}=\sqrt{6};\)
\(ab=\dfrac{\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\right)}{2\cdot2}=\dfrac{6-2}{4}=\dfrac{4}{4}=1\)
Ta có:
\(B=\dfrac{1}{a^7}+\dfrac{1}{b^7}=\dfrac{a^7+b^7}{\left(ab\right)^7}=a^7+b^7\)
\(=\left(a^3+b^3\right)\left(a^4+b^4\right)-a^3b^4-a^4b^3=\left[\left(a+b\right)^3-3ab\left(a+b\right)\right]\left[\left(a^2+b^2\right)^2-2a^2b^2\right]-a^3b^3\left(a+b\right)=\left[\left(a+b\right)^3-3ab\left(a+b\right)\right]\left\{\left[\left(a+b\right)^2-2ab\right]^2-2a^2b^2\right\}-a^3b^3\left(a+b\right)=\left[6\sqrt{6}-3\sqrt{6}\right]\left[16-2\right]-\sqrt{6}=3\sqrt{6}\cdot14-\sqrt{6}=41\sqrt{6}\)
