Giải câu 1 trang 60 toán VNEN 8 tập 1

Làm các bài tập tự luận

Câu 1: Trang 60 toán VNEN 8 tập 1

Thực hiện các phép tính sau:

a) ( - $\frac{2x–1}{2x+1}$) : $\frac{4x}{10x–5}$; b) ($\frac{1}{x^2+1}$ - $\frac{2–x}{x+1}$) : ($\frac{1}{x}$ + x – 2);

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

d) ( + $\frac{y}{x^2+y^2}$) : ($\frac{1}{x–y}$ - $\frac{2xy}{x^3-x^{2}y+x{y}^2–y^3}$).

Bài làm:

a) ( - $\frac{2x–1}{2x+1}$) : $\frac{4x}{10x–5}$ = $\frac{(2x+1{)}^2–(2x–1{)}^2}{(2x–1)(2x+1)}$ : $\frac{4x}{5(2x–1)}$ = $\frac{8x}{(2x–1)(2x+1)}$ . $\frac{5(2x–1)}{4x}$ = $\frac{10}{2x+1}$;

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

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

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

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

d) ( + $\frac{y}{x^2+y^2}$) : ($\frac{1}{x–y}$ - $\frac{2xy}{x^3-x^{2}y+x{y}^2–y^3}$) = ($\frac{x(x+y)}{(x^2+y^2)(x+y)}$ + $\frac{y}{x^2+y^2}$) : ($\frac{1}{x-y}$ - $\frac{2xy}{(x–y)(x^2+y^2)}$)

= : $\frac{(x–y{)}^2}{(x–y)(x^2+y^2)}$ = . $\frac{x^2+y^2}{x-y}$ = $\frac{x+y}{x-y}$.