bn viết rõ đề đc ko

tc tỉ lệ thức,ta có (ra số hơi to thông cảm )

\(\Leftrightarrow\frac{22990993x+80556557}{1870865100}=-\frac{4}{1}\Rightarrow\left(22990993x+80556557\right)1=-1870865100.4\)

\(\frac{\left(22990993x+80556557\right)1}{22990933x}=-\frac{1870865100.4}{22990993x}\)( chia 2 vế cho 2290933x)

\(\Rightarrow\frac{22990933x+80556557}{22990933x}=-\frac{1870865100.4}{22990933x}\)

=>x=-329

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

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

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

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

Vì \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}\ne0\Rightarrow x+329=0\Leftrightarrow x=-329\)

Sửa đề: Tìm x \(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)

Bài giải

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

\(\Leftrightarrow\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+329+20}{5}=0\)

\(\Leftrightarrow\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+329}{5}+\dfrac{20}{5}=0\)

\(\Leftrightarrow\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+329}{5}+4=0\)

\(\Leftrightarrow\dfrac{x+2}{327}+1+\dfrac{x+3}{326}+1+\dfrac{x+4}{325}+1+\dfrac{x+5}{324}+1+\dfrac{x+329}{5}=0\)

\(\Leftrightarrow\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)

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

\(\Rightarrow x+329=0\). Do \(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\ne0\)

\(\Rightarrow x=-329\)