Quan sát hình vẽ và cho biết tia đối của tia \[AB\] là tia 

Ta có \(F = \frac{1}{{18}} + \frac{1}{{54}} + \frac{1}{{108}} + ... + \frac{1}{{990}}\)

\( = \frac{1}{{3.6}} + \frac{1}{{6.9}} + \frac{1}{{9.12}} + ... + \frac{1}{{30.33}}\)

\( = \frac{1}{3}.\left( {\frac{3}{{3.6}} + \frac{3}{{6.9}} + \frac{3}{{9.12}} + ... + \frac{3}{{30.33}}} \right)\)

\( = \frac{1}{3}.\left( {\frac{{6 - 3}}{{3.6}} + \frac{{9 - 6}}{{6.9}} + \frac{{12 - 9}}{{9.12}} + ... + \frac{{33 - 30}}{{30.33}}} \right)\)

\( = \frac{1}{3}.\left[ {\left( {\frac{6}{{3.6}} - \frac{3}{{3.6}}} \right) + \left( {\frac{9}{{6.9}} - \frac{6}{{6.9}}} \right) + \left( {\frac{{12}}{{9.12}} - \frac{9}{{9.12}}} \right) + ... + \left( {\frac{{33}}{{30.33}} - \frac{{30}}{{30.33}}} \right)} \right]\)

\( = \frac{1}{3}.\left( {\frac{1}{3} - \frac{1}{6} + \frac{1}{6} - \frac{1}{9} + \frac{1}{9} - \frac{1}{{12}} + ... + \frac{1}{{30}} - \frac{1}{{33}}} \right)\)

\( = \frac{1}{3}.\left( {\frac{1}{3} - \frac{1}{{33}}} \right)\)\( = \frac{1}{3}.\frac{{10}}{{33}} = \frac{{10}}{{99}}\).

Vậy \(F = \frac{{10}}{{99}}\).

Đáp án đúng là: D

Ta có \(\frac{x}{9} = \frac{{ - \,4}}{3}\).

Suy ra \(x = \frac{{ - \,4\,\,.\,\,9}}{3} =  - \,12\).