ĐK: \(x\ne\pm3\)
Khi đó:
\(C=\dfrac{2\left(x-3\right)}{x^2-9}+\dfrac{1\left(x+3\right)}{x^2-9}-\dfrac{8}{x^2-9}\\ =\dfrac{2x-6}{x^2-9}+\dfrac{x+3}{x^2-9}-\dfrac{8}{x^2-9}\\ =\dfrac{2x-6+x+3-8}{x^2-9}\\ =\dfrac{3x-11}{x^2-9}\)
Thế x = 4 vào C được:
\(C=\dfrac{3.4-11}{4^2-9}=\dfrac{12-11}{16-9}=\dfrac{1}{7}\)