Tìm n biết:

C

- Hướng dẫn giải

Đáp án C

Ta có: $\frac{1}{{\log }_{2}x}+\frac{1}{{\log }_{2^2}x}+\frac{1}{{\log }_{2^3}x}+\mathrm{...}+\frac{1}{{\log }_{2^n}x}$

$={\log }_{x}2+{\log }_x{2}^2+{\log }_x{2}^3+\mathrm{...}+{\log }_x{2}^n$ 

$={\log }_x\left({2.2}^2{.2}^3{\mathrm{...2}}^n\right)=465{\log }_{x}2={\log }_x{2}^{465}$

$$\begin{gathered} \Rightarrow {2.2}^2{.2}^3{\mathrm{...2}}^n\iff 1+2+3+\mathrm{...}+n=465 \\ \iff \frac{n}{2}\left(n+1\right)=465\iff n^2+n-930=0 \\ \iff \left[\begin{array}{l}n=30\\ n=-31\end{array}\right.\Rightarrow n=30. \end{gathered}$$