Giải câu 2 trang 46 toán VNEN 8 tập 1.
a) $\frac{2}{x + 3}$ + $\frac{1}{x}$ = $\frac{2x}{x(x + 3)}$ + $\frac{x + 3}{x(x + 3)}$ = $\frac{2x + x + 3}{x(x + 3)}$ = $\frac{3x + 3}{x(x + 3)}$;
b) $\frac{x + 1}{2x - 2}$ + $\frac{-2x}{x^{2} - 1}$ = $\frac{x + 1}{2(x – 1)}$ + $\frac{-2x}{(x – 1)(x + 1)}$ = $\frac{(x + 1)(x + 1)}{2(x – 1)(x + 1)}$ + $\frac{2.(-2x)}{2(x – 1)(x + 1)}$ = $\frac{x^{2} + 2x + 1 – 4x}{2(x – 1)(x + 1)}$ = $\frac{x^{2} – 2x + 1 }{2(x – 1)(x + 1)}$ = $\frac{x - 1}{2(x + 1)}$;
c) $\frac{y - 12}{6y - 36}$ + $\frac{4}{y^{2} – 6y}$ = $\frac{y - 12}{6(y – 6)}$ + $\frac{4}{y(y – 6)}$ = $\frac{y(y – 12)}{6y(y – 6)}$ + $\frac{6.4}{6y(y – 6)}$ = $\frac{y^{2} – 12y + 24}{6y(y – 6)}$;
d) $\frac{6 - x}{x^{2} + 3x}$ + $\frac{3}{2x + 6}$ = $\frac{6 - x}{x(x + 3)}$ + $\frac{3}{2(x + 3)}$ = $\frac{2(6 – x)}{2x(x + 3)}$ + $\frac{3.x}{2x(x + 3)}$ = $\frac{12 – 2x + 3x}{2x(x + 3)}$ = $\frac{x + 12}{2x(x + 3)}$.