Đáp án:$\frac{{2017}}{{x(x + 2017)}}$
Giải thích các bước giải:
$\eqalign{
& P = \frac{1}{{x(x + 1)}} + \frac{1}{{(x + 1)(x + 2)}} + ... + \frac{1}{{(x + 2016)(x + 2017)}} \cr
& = \frac{{(x + 1) - x}}{{x(x + 1)}} + \frac{{(x + 2) - (x + 1)}}{{(x + 1)(x + 2)}} + ... + \frac{{(x + 2017) - (x + 2016)}}{{(x + 2016)(x + 2017)}} \cr
& = \frac{1}{x} - \frac{1}{{x + 1}} + \frac{1}{{x + 1}} - \frac{1}{{x + 2}} + ... + \frac{1}{{x + 2016}} - \frac{1}{{x + 2017}} \cr
& = \frac{1}{x} - \frac{1}{{x + 2017}} \cr
& = \frac{{(x + 2017) - x}}{{x(x + 2017)}} \cr
& = \frac{{2017}}{{x(x + 2017)}} \cr} $
