Giải câu 2 trang 46 toán VNEN 8 tập 1

Câu 2: Trang 46 toán VNEN 8 tập 1

Thực hiện các phép tính sau:

a) + $\frac{1}{x}$; b) $\frac{x+1}{2x-2}$ + $\frac{-2x}{x^2-1}$;

c) + $\frac{4}{y^2–6y}$; d) $\frac{6-x}{x^2+3x}$ + $\frac{3}{2x+6}$.

Bài làm:

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

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

c) + $\frac{4}{y^2–6y}$ = $\frac{y-12}{6(y–6)}$ + $\frac{4}{y(y–6)}$ = $\frac{y(y–12)}{6y(y–6)}$ + $\frac{6.4}{6y(y–6)}$ = $\frac{y^2–12y+24}{6y(y–6)}$;

d) + $\frac{3}{2x+6}$ = $\frac{6-x}{x(x+3)}$ + $\frac{3}{2(x+3)}$ = $\frac{2(6–x)}{2x(x+3)}$ + $\frac{3.x}{2x(x+3)}$ = $\frac{12–2x+3x}{2x(x+3)}$ = $\frac{x+12}{2x(x+3)}$.