[กิตติ]

$2x=1+\sqrt{5} $
$2x-1=\sqrt{5}$
$(2x-1)^2=5 \rightarrow 4x^2-4x+1=5$
$x^2-x-1=0 \rightarrow x-1-\frac{1}{x} =0$
$x^2-x=1$
$x-\frac{1}{x}=1 \rightarrow x^2+\frac{1}{x^2}=3$
$x^3-\frac{1}{x^3}=(x-\frac{1}{x})(x^2+\frac{1}{x^2}+1)=4$

$(x^3-\frac{1}{x^3})(x^2+\frac{1}{x^2})=12$
$x^5-\frac{1}{x^5}+x-\frac{1}{x}=4$
$x^5-\frac{1}{x^5}=11$

$(x^5-\frac{1}{x^5})(x^2+\frac{1}{x^2})=33$
$x^7-\frac{1}{x^7}+x^3-\frac{1}{x^3}=33$
$x^7-\frac{1}{x^7}=29$

ตอบ D