\(C=\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{2x\left(x-1\right)}{\left(x+3\right)\left(x-3\right)}\) ( x khác 3 ; -3 )
\(=\frac{x^2+4x+3+x^2-4x+3+2x^2-2x}{\left(x+3\right)\left(x-3\right)}=\frac{4x^2-2x+6}{\left(x+3\right)\left(x-3\right)}=\frac{2\left(2x^2-x+3\right)}{\left(x+3\right)\left(x-3\right)}\)
Tại x = 5=> \(C=\frac{2\left(2.5^2-5+3\right)}{\left(5+3\right)\left(5-3\right)}=\frac{48}{8}=6\)