đề rối
\(x+\frac{3}{10}+x+\frac{3}{11}+x+\frac{3}{12}=x+\frac{3}{13}+x+\frac{3}{14}\)
\(\Leftrightarrow3x+\left(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}\right)=2x+\left(\frac{3}{13}+\frac{3}{14}\right)\)
\(\Leftrightarrow3x+\frac{181}{220}=2x+\frac{81}{182}\)
\(\Leftrightarrow3x-2x=-\frac{181}{220}+\frac{81}{182}\)
\(\Leftrightarrow x=-\frac{181}{220}+\frac{81}{182}=\frac{-7561}{20020}\)
hay là
\(\frac{x+3}{10}+\frac{x+3}{11}+\frac{x+3}{12}=\frac{x+3}{13}+\frac{x+3}{14}\)
\(\Leftrightarrow\frac{x+3}{10}+\frac{x+3}{11}+\frac{x+3}{12}-\frac{x+3}{13}+\frac{x+3}{14}=0\)
\(\Leftrightarrow\left(x+3\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+3=0\)( vì \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)\ne0\))
\(\Rightarrow x=-3\)
