\(\frac{a^2-5ab+4}{16-a^2}-\frac{2a}{2a^2+8a}\)
\(=\frac{a^2-5a+4}{\left(4-a\right)\left(4+a\right)}-\frac{2a}{2a\left(a+4\right)}\)
\(=\frac{a^2-5a+4-\left(4-a\right)}{\left(4-a\right)\left(4+a\right)}\)
\(=\frac{a^2-4a}{\left(4-a\right)\left(4+a\right)}=\frac{a\left(a-4\right)}{\left(4-a\right)\left(4+a\right)}=\frac{-a}{4+a}\)
PS:Quy đồng sai chỗ nào tự coi lại nhá