Question

tìm x biết \(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}\)=-4 

Answer 1

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

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

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

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

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

\(x=0-329=-329\)