Giải bài tập 3 trang 42 sgk Toán 8 tập 1 CD

a. $\frac{1}{x-2}-\frac{1}{x+1}$ = $\frac{x+1-x+2}{(x-2)(x+1)}=\frac{3}{(x-2)(x+1)}$

b. $\frac{12}{x^{2}-9}-\frac{2}{x-3}$ = $\frac{12}{(x-3)(x+3)}-\frac{2(x+3)}{(x-3)(x+3)}=\frac{12-2x-6}{(x-3)(x+3)}$

= $\frac{6-2x}{(x-3)(x+3)}=\frac{-2(x-3)}{(x-3)(x+3)}=\frac{-2}{x+3}$

c. $\frac{1}{xy-x^{2}}-\frac{1}{y^{2}-xy}$ = $\frac{1}{x(y-x)}-\frac{1}{y(y-x)}=\frac{y}{x(y-x)y}-\frac{x}{y(y-x)x}=\frac{y-x}{(y-x)xy}=\frac{1}{xy}$

d. $\frac{2x}{x^{2}-1}-\frac{3}{2+2x}+\frac{1}{2+2x}$

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

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

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

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

= $\frac{4x-2x+2}{(x-1)(x+1).2}$

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

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