Đáp án:
Giải thích các bước giải:
a) $\frac{3}{x+3}+\frac{1}{x-3}-\frac{18}{9-x^2}\\=\frac{4x-6}{(x-3)(x+3)}+\frac{18}{x^2-9}\\=\frac{4(x-3)}{(x-3)(x+3)}\\=\frac{4}{(x+3)}$
b) $\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6}{1-x}\\=\frac{4x^2-3x+17}{(x-1)(x^2+x+1)}+\frac{2x-1}{x^2+x+1}-\frac{6}{x-1}\\=\frac{4x^2-3x+17}{(x-1)(x^2+x+1)}+\frac{-4x^2-7x-5}{(x-1)(x^2+x+1)}\\=\frac{-10x+12}{(x-1)(x^2+x+1)}$
c) $\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\\=\frac{x+2}{x+3}-\frac{1}{x-2}-\frac{5}{(x+3)(x-2)}\\=\frac{x^2+x-9}{(x+3)(x-2)}$