Question

Tìm x :

\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)

Answer 1

\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}\)\(+\frac{x+349}{5}=0\)

=>  \(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)\)\(+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)\)\(+\left(\frac{x+349}{5}-4\right)=0\)

=>   \(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}\)\(+\frac{x+329}{324}+\frac{x+329}{5}=0\)

=> \(\left(x+329\right)\)\(\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\)\(=0\)

=>  \(x+329=0\)\(\left(do\frac{1}{327}+...+\frac{1}{324}+\frac{1}{5}\ne0\right)\)

=> \(x=-329\)

Vậy x = -329