\(\frac{1}{3}\sqrt{9+6a+a^2}+\frac{4a}{3}+5\)
\(=\frac{1}{3}\sqrt{\left(a+3\right)^2}+\frac{4a}{3}+5\)
\(=\frac{1}{3}\left|a+3\right|+\frac{4a}{3}+5\)(1)
Với a < 3 \(\left(1\right)=-\frac{1}{3}\left(a+3\right)+\frac{4}{3}a+5=a+4\)
Với a >= 3 \(\left(1\right)=\frac{1}{3}\left(a+3\right)+\frac{4}{3}a+5=\frac{5}{3}a+6\)