$\leftrightarrow \dfrac{x+2}{327} +1+ \dfrac{x+3}{326} +1+ \dfrac{x+4}{325} +1+\dfrac{x+5}{324} +1+ \dfrac{x+349}{5} -4 = 0$
$\leftrightarrow \dfrac{x+ 329}{327} + \dfrac{x+ 329}{326} + \dfrac{x+ 329}{325}+ \dfrac{x+ 329}{324} + \dfrac{x+ 329}{5} =0$
$\leftrightarrow (x+ 329).(\dfrac{1}{327} + \dfrac{1}{326} + \dfrac{1}{325} + \dfrac{1}{324}+\dfrac{1}{5}) =0$
Do $( \dfrac{1}{327} +\dfrac{1}{326} + \dfrac{1}{325} + \dfrac{1}{324} +\dfrac{1}{5}) >0$ nên $x+ 329 =0 \rightarrow x= -329$