Question

Tính giá trị S = \(\dfrac{1}{a^7}+\dfrac{1}{b^7}\) với a = \(\dfrac{\sqrt{6}+\sqrt{2}}{2}\);b= \(\dfrac{\sqrt{6}-\sqrt{2}}{2}\)

Answer 1

Lời giải:
$a+b=\frac{\sqrt{6}+\sqrt{2}+\sqrt{6}-\sqrt{2}}{2}=\sqrt{6}$

$ab=\frac{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})}{2.2}=\frac{6-2}{4}=1$

Khi đó:
$S=\frac{1}{a^7}+\frac{1}{b^7}=\frac{a^7+b^7}{a^7b^7}$

$=\frac{a^7+b^7}{(ab)^7}=\frac{a^7+b^7}{1}=a^7+b^7$

$=(a^3+b^3)(a^4+b^4)-a^3b^3(a+b)$

$=(a^3+b^3)(a^4+b^4)-(a+b)$

Ta có:

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

$a^4+b^4=(a^2+b^2)^2-2a^2b^2=(a^2+b^2)^2-2$

$=[(a+b)^2-2ab]^2-2=(6-2)^2-2=14$

$S=3\sqrt{6}.14-\sqrt{6}=41\sqrt{6}$