\(\dfrac{1}{4}+\dfrac{1}{3}:\left(2x+1\right)=-5\)
\(\Rightarrow\dfrac{1}{3}:\left(2x+1\right)=-5-\dfrac{1}{4}\)
\(\Rightarrow\dfrac{1}{3}:\left(2x+1\right)=\dfrac{-20}{4}-\dfrac{1}{4}\)
\(\Rightarrow\dfrac{1}{3}:\left(2x+1\right)=\dfrac{-21}{4}\)
\(\Rightarrow2x+1=\dfrac{1}{3}:\dfrac{-21}{4}\)
\(\Rightarrow2x+1=\dfrac{1}{3}.\dfrac{-4}{21}\)
\(\Rightarrow2x+1=\dfrac{-4}{63}\)
\(\Rightarrow2x=\dfrac{-4}{63}-1\)
\(\Rightarrow2x=\dfrac{-4}{63}-\dfrac{63}{63}\)
\(\Rightarrow2x=\dfrac{-67}{63}\)
\(\Rightarrow x=\dfrac{-67}{63}:2\)
\(\Rightarrow x=\dfrac{-67}{63}.\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{-67}{126}\)
Vậy \(x=\dfrac{-67}{126}\)
