Đáp án: $\dfrac{1}{x-1}$
Giải thích các bước giải:
$\dfrac{8}{(x^{2}+3)(x+1)(x-1)}+\dfrac{2}{x^{2}+3}+\dfrac{1}{x+1}\\=\dfrac{8}{(x^{2}+3)(x^{2}-1)}+\dfrac{2}{x^{2}+3}+\dfrac{1}{x+1}\\=\dfrac{8}{(x^{2}+3)(x-1)(x+1)}+\dfrac{2}{x^{2}+3}+\dfrac{1}{x+1}\\=\dfrac{8+2(x-1)(x+1)+(x^{2}+3)(x-1)}{(x^{2}+3)(x-1)(x+1)}\\=\dfrac{8+2(x^{2}-1)+x^{3}-x^{2}+3x-3}{(x^{2}+3)(x-1)(x+1)}\\=\dfrac{8+2x^{2}-2+x^{3}-x^{2}+3x-3}{(x^{2}+3)(x-1)(x+1)}\\=\dfrac{3+x^{2}+x^{3}+3x}{(x^{2}+3)(x-1)(x+1)}\\=\dfrac{3+x^{2}+x(x^{2}+3)}{(x^{2}+3)(x-1)(x+1)}\\=\dfrac{(x^{2}+3)(1+x)}{(x^{2}+3)(x-1)(x+1)}\\=\dfrac{1}{x-1}$
