Tìm x của biểu thức : b) 1/ 1.2 + 1 /2.3 + 1/ 3.4 + . . . + 1 /x ( x + 1 ) = 2008 /2009 ;

b) \[\frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{x\left( {x + 1} \right)}} = \frac{{2008}}{{2009}}\]

\[\frac{{2 - 1}}{{1.2}} + \frac{{3 - 2}}{{2.3}} + \frac{{4 - 3}}{{3.4}} + ... + \frac{{\left( {x + 1} \right) - x}}{{x\left( {x + 1} \right)}} = \frac{{2008}}{{2009}}\]

\[\frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{x} - \frac{1}{{x + 1}} = \frac{{2008}}{{2009}}\]

\[\frac{1}{1} - \frac{1}{{x + 1}} = \frac{{2008}}{{2009}}\]

\[\frac{x}{{x + 1}} = \frac{{2008}}{{2009}}\]

\[\frac{{2009x - 2008\left( {x + 1} \right)}}{{2009\left( {x + 1} \right)}} = 0\]

\[2009x - 2008x - 2008 = 0\]

\[x - 2008 = 0\]

\[x = 2008\].

Vậy \[x = 2008\].