<=>\(\dfrac{\left(x-3\right)9}{99}+\dfrac{\left(x+1\right)33}{99}=\dfrac{\left(x+7\right)11}{99}-\dfrac{99}{99}\)
=>(x-3)9 + (x+1)33 = (x+7)11 - 99
=>9x-27+33x+33=11x+77-99
=>9x+33x-11x=27-33+77-99
=>31x=-28
=>x=\(\dfrac{-28}{31}\)
\(\dfrac{x-3}{11}+\dfrac{x+1}{3}=\dfrac{x+7}{9}-1\)
\(\Leftrightarrow\dfrac{9\left(x-3\right)+33\left(x+1\right)}{99}=\dfrac{11\left(x+7\right)-99}{99}\)
\(\Leftrightarrow9\left(x-3\right)+33\left(x+1\right)=11\left(x+7\right)-99\)
\(\Leftrightarrow9x-27+33x+33=11x+77-99\)
\(\Leftrightarrow9x+33x-11x=77-99+27-33\)
\(\Leftrightarrow31x=-28\)
\(\Leftrightarrow x=-\dfrac{28}{31}\)
Vậy phương trình có nghiệm duy nhất \(x=-\dfrac{28}{31}\)
