Cho hai số thực a,b thỏa mãn đi...

A

- Hướng dẫn giải

Chọn A.

Ta có:

$$\begin{gathered} \frac{1}{{\log }_{\mathrm{a}}\mathrm{b}}-\frac{1}{{\log }_{\mathrm{b}}\mathrm{a}}=\sqrt{2018} \\ \iff {\log }_{\mathrm{a}}\mathrm{b}+\frac{1}{{\log }_{\mathrm{a}}\mathrm{b}}=\sqrt{2018} \\ \iff \mathrm{t}+\frac{1}{\mathrm{t}}=\sqrt{2018} \end{gathered}$$ 

 $$\begin{gathered} \mathrm{P}=\frac{1}{{\log }_{\mathrm{ab}}\mathrm{b}}-\frac{1}{{\log }_{\mathrm{ab}}\mathrm{a}}={\log }_{\mathrm{b}}\mathrm{ab}-{\log }_{\mathrm{a}}\mathrm{ab} \\ ={\log }_{\mathrm{b}}\mathrm{a}-{\log }_{\mathrm{a}}\mathrm{b}=\frac{1}{{\log }_{\mathrm{a}}\mathrm{b}}-{\log }_{\mathrm{a}}\mathrm{b}=\frac{1}{\mathrm{t}}-t \end{gathered}$$

Mà $(\mathrm{t}+\frac{1}{\mathrm{t}}{)}^2-(\frac{1}{\mathrm{t}}-\mathrm{t}{)}^2=4$ suy ra

$$\begin{gathered} \mathrm{P}=\frac{1}{\mathrm{t}}-\mathrm{t}=\sqrt{(\mathrm{t}+\frac{1}{\mathrm{t}}{)}^2-4} \\ =\sqrt{2018-4}=\sqrt{2014}. \end{gathered}$$