Đáp án:

$ P =\dfrac{6049}{4034}  $

Lời giải:

$\eqalign{ & f'\left( x \right) + \left( {2x + 3} \right){f^2}\left( x \right) = 0 \cr & \Leftrightarrow {{f'\left( x \right)} \over {{f^2}\left( x \right)}} = - \left( {2x + 3} \right) \cr & \Leftrightarrow \int\limits_{}^{} {{{f'\left( x \right)} \over {{f^2}\left( x \right)}}dx} = - \int\limits_{}^{} {\left( {2x + 3} \right)dx} \cr & \Leftrightarrow - {1 \over {f\left( x \right)}} = - \left( {{x^2} + 3x} \right) + C \cr & \text{Thay }x = 1 \cr & \Rightarrow - {1 \over {f\left( 1 \right)}} = - \left( {{1^2} + 3.1} \right) + C \cr & \Leftrightarrow - {1 \over {1/6}} = - 4 + C \cr & \Leftrightarrow - 6 = - 4 + C \Leftrightarrow C = - 2 \cr & \Rightarrow - {1 \over {f\left( x \right)}} = - \left( {{x^2} + 3x} \right) - 2 \cr & \Leftrightarrow f\left( x \right) = {1 \over {{x^2} + 3x + 2}} = {1 \over {\left( {x + 1} \right)\left( {x + 2} \right)}} = {1 \over {x + 1}} - {1 \over {x + 2}} \cr & P = 1 + f\left( 1 \right) + f\left( 2 \right) + ... + f\left( {2017} \right) \cr & P = 1 + {1 \over 2} - {1 \over 3} + {1 \over 3} - {1 \over 4} + ... + {1 \over {2016}} - {1 \over {2017}} \cr & P = 1 + {1 \over 2} - {1 \over {2017}} \cr & P = {{6049} \over {4034}} \cr} $