Giải bài tập 9 trang 36 SBT toán 8 tập 1 cánh diều:

a) $\frac{x+2y}{a}+\frac{x-2y}{a}=\frac{x+2y+x-2y}{a}=\frac{2x}{a}$.

b) $\frac{x}{x-1}+\frac{1}{1-x}=\frac{x}{x-1}+\frac{-1}{x-1}=\frac{x-1}{x-1}=1$.

c) $ \frac{x^{2}+2}{x^{3}-1}+\frac{2}{x^{2}+x+1}+\frac{1}{1-x}$

= $\frac{x^{2}+2}{(x-1)(x^{2}+x+1)}+\frac{2(x-1)}{(x^{2}+x+1)(x-1)}+\frac{-1(x^{2}+x+1)}{(x-1)(x^{2}+x+1)}$

= $\frac{x^{2}+2+2(x-1)-(x^{2}+x+1)}{(x-1)(x^{2}+x+1)}$ 

= $\frac{x-1}{(x-1)(x^{2}+x+1)}$

= $\frac{1}{x^{2}+x+1}$

d) $x+\frac{1}{x+1}-1=\frac{x(x+1)}{x+1}+\frac{1}{x+1}-\frac{x+1}{x+1}$

= $\frac{x(x+1)+1-(x+1)}{x+1}$

= $\frac{x^{2}+x+1-x-1}{x+1}$

= $\frac{x^{2}}{x+1}$