ĐKXĐ: \(a\ne-1;a\ne-5\)
\(\frac{a}{a+1}-\frac{1}{a+5}=\frac{1}{3}\)
\(\Leftrightarrow\frac{3a\left(a+5\right)-3a\left(a+1\right)}{3\left(a+1\right)\left(a+5\right)}=\frac{\left(a+1\right)\left(a+5\right)}{3\left(a+1\right)\left(a+5\right)}\)
\(\Rightarrow3a^2+15a-3a^2-3a=a^2+5a+a+5\)
\(\Leftrightarrow12a=a^2+6a+5\)
\(\Leftrightarrow a^2-6a+5=0\)
\(\Leftrightarrow a^2-a-5a+5=0\)
\(\Leftrightarrow a\left(a-1\right)-5\left(a-1\right)=0\)
\(\Leftrightarrow\left(a-1\right)\left(a-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a-1=0\\a-5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a=1\left(TM\right)\\a=5\left(TM\right)\end{matrix}\right.\)
Vậy a ={1; 5}
