$\displaystyle \begin{array}{{>{\displaystyle}l}} \frac{x+2}{327} +\frac{x+3}{326} +\frac{x+4}{325} +\frac{x+5}{324} +\frac{x+349}{5} =0\ \\ \frac{x+2}{327} +1+\frac{x+3}{326} +1+\frac{x+4}{325} +1+\frac{x+5}{324} +1+\frac{x+349}{5} -4=0\ \\ \frac{x+329}{327} +\frac{x+329}{326} +\frac{x+329}{325} +\frac{x+329}{324} +\frac{x+329}{5} =0\ \\ ( x+329)\left(\frac{1}{327} +\frac{1}{326} +\frac{1}{325} +\frac{1}{324} +\frac{1}{5}\right) =0\ \\ \rightarrow x+329=0\ \rightarrow x=-329\ \\ Vì\ \left(\frac{1}{327} +\frac{1}{326} +\frac{1}{325} +\frac{1}{324} +\frac{1}{5} \#0\right) \ \\ b)( x-1)^{5} =-243\\ ( x-1)^{5} =( -3)^{5}\\ \rightarrow x-1=-3\\ \rightarrow x=-2\ \\ c)\frac{x+2}{11} +\frac{x+2}{12} +\frac{x+2}{13} +=\frac{x+2}{14} +\frac{x+2}{15} \ \\ ( x+1)\left(\frac{1}{11} +\frac{1}{12} +\frac{1}{13} -\frac{1}{14} -\frac{1}{15}\right) =0\ \\ \rightarrow ( x+1) =0\rightarrow x=-1\ \\ Vì\ \left(\frac{1}{11} +\frac{1}{12} +\frac{1}{13} -\frac{1}{14} -\frac{1}{15} \#0\right) \ \\ d) 2|5x-3|-2x=14\ \\ \rightarrow |5x-3|-x=7\\ \rightarrow \left[ \begin{array}{l l} 5x-3-x=7\left( x\geqslant \frac{3}{5}\right) & \\ 3-5x-x=7\left( x< \frac{3}{5}\right) & \end{array} \right.\\ \rightarrow \left[ \begin{array}{l l} x=\frac{5}{2} & \\ x=-\frac{2}{3} & \end{array} \right.\\ \end{array}$
