Giải câu 3 trang 48 toán VNEN 8 tập 1

a) $\frac{3}{2y + 4}$ - $\frac{1}{3y + 6}$ = $\frac{3}{2(y + 2)}$ - $\frac{1}{3(y + 2)}$ = $\frac{3.3}{6(y + 2)}$ - $\frac{2}{6(y + 2)}$ = $\frac{9 - 2}{6(y + 2)}$ = $\frac{7}{6(y + 2)}$;

b) $\frac{1}{2x - 3}$ - $\frac{1}{2x + 3}$ = $\frac{2x + 3}{(2x – 3)(2x + 3)}$ - $\frac{2x - 3}{(2x + 3)(2x – 3)}$ = $\frac{2x + 3 – (2x – 3)}{(2x + 3)(2x – 3)}$ = $\frac{6}{(2x + 3)(2x – 3)}$;             

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

d) $\frac{x + 1}{x + 4}$ - $\frac{x^{2} - 4}{x^{2} - 16}$ = $\frac{x + 1}{x + 4}$ - $\frac{x^{2} - 4}{(x + 4)(x – 4)}$

= $\frac{(x + 1)(x – 4)}{(x + 4)(x – 4)}$ - $\frac{x^{2} - 4}{(x + 4)(x – 4)}$ = $\frac{x^{2} – 3x – 4 - x^{2} + 4}{(x + 4)(x – 4)}$ = $\frac{-3x}{(x – 4)(x + 4)}$.