Đáp án:
\[\frac{{27}}{{28}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
f'\left( x \right) = \left( {2x + 1} \right).{f^2}\left( x \right)\\
\Leftrightarrow \frac{{f'\left( x \right)}}{{{f^2}\left( x \right)}} = 2x + 1\\
\Leftrightarrow \int {\frac{{f'\left( x \right)}}{{{f^2}\left( x \right)}}dx} = \int {\left( {2x + 1} \right)dx} \\
\Leftrightarrow - \frac{1}{{f\left( x \right)}} = {x^2} + x + C\\
\Rightarrow f\left( x \right) = \frac{{ - 1}}{{{x^2} + x + C}}\\
f\left( 1 \right) = - 0,5 \Leftrightarrow \frac{{ - 1}}{{{1^2} + 1 + C}} = - \frac{1}{2} \Leftrightarrow C = 0\\
\Rightarrow f\left( x \right) = \frac{1}{{{x^2} + x}} = \frac{1}{{x\left( {x + 1} \right)}} = \frac{1}{x} - \frac{1}{{x + 1}}\\
f\left( 1 \right) + f\left( 2 \right) + .... + f\left( {27} \right)\\
= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ..... + \frac{1}{{27}} - \frac{1}{{28}}\\
= 1 - \frac{1}{{28}} = \frac{{27}}{{28}}
\end{array}\)
