- \(ĐKXĐ:\left\{{}\begin{matrix}a+2\ne0\\a\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a\ne-2\\a\ne0\end{matrix}\right.\)
- Ta có : \(\dfrac{3a}{a+2}+\dfrac{a-2}{a}=0\)
\(\Leftrightarrow\dfrac{3a^2}{a\left(a+2\right)}+\dfrac{\left(a-2\right)\left(a+2\right)}{a\left(a+2\right)}=0\)
\(\Leftrightarrow\dfrac{3a^2+\left(a-2\right)\left(a-2\right)}{a\left(a+2\right)}=0\)
\(\Leftrightarrow\dfrac{3a^2+a^2-4}{a\left(a+2\right)}=0\Leftrightarrow4a^2-4=0\)
\(\Leftrightarrow4\left(a^2-1\right)=0\Leftrightarrow4\left(a-1\right)\left(a+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a-1=0\\a+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=1\\a=-1\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = {-1; 1 }
