Hướng dẫn Giải Bài 25 (Trang 47, SGK Toán 8, Tập 1)

Làm tính cộng các phân thức sau:

a) $\frac{5}{2 x^{2} y} + \frac{3}{5 x y^{2}} + \frac{x}{y^{3}};$ b) $\frac{x + 1}{2 x + 6} + \frac{2 x + 3}{x (x + 3)};$

c) $\frac{3 x + 5}{x^{2} - 5 x} + \frac{25 - x}{25 - 5 x}$ ; d) $x^{2} + \frac{x^{4} + 1}{1 - x^{2}} + 1$ ;

e) $\frac{4 x^{2} - 3 x + 17}{x^{3} - 1} + \frac{2 x - 1}{x^{2} + x + 1} + \frac{6}{1 - x} .$

Giải

a) $\frac{5}{2 x^{2} y} + \frac{3}{5 x y^{2}} + \frac{x}{y^{3}} = \frac{5.5 y^{2}}{2 x^{2} y . 5 y^{2}} + \frac{3.2 x y}{5 x y^{2} . 2 x y} + \frac{x . 10 x^{2}}{y^{3} . 10 x^{2}} = \frac{25 y^{2}}{10 x^{2} y^{3}} + \frac{6 x y}{10 x^{2} y^{3}} + \frac{10 x^{3}}{10 x^{2} y^{3}} = \frac{25 y^{2} + 6 x y + 10 x^{3}}{10 x^{2} y^{3}}$

b) $\frac{x + 1}{2 x + 6} + \frac{2 x + 3}{x (x + 3)} = \frac{x + 1}{2 (x + 3)} + \frac{2 x + 3}{x (x + 3)} = \frac{x (x + 1)}{2 x (x + 3)} + \frac{2 (2 x + 3)}{2 x (x + 3)} = \frac{x^{2} + x + 4 x + 6}{2 x (x + 3)} = \frac{x^{2} + 5 x + 6}{2 x (x + 3)} = \frac{x^{2} + 2 x + 3 x + 6}{2 x (x + 3)} = \frac{x (x + 2) + 3 (x + 2)}{2 x (x + 3)} = \frac{(x + 2) (x + 3)}{2 x (x + 3)} = \frac{x + 2}{2 x}$

c) $\frac{3 x + 5}{x^{2} - 5 x} + \frac{25 - x}{25 - 5 x} = \frac{3 x + 5}{x (x - 5)} + \frac{25 - x}{5 (5 - x)} = \frac{3 x + 5}{x (x - 5)} + \frac{x - 25}{5 (x - 5)} = \frac{5 . (3 x + 5)}{5 x (x - 5)} + \frac{x . (x - 25)}{5 x (x - 5)} = \frac{15 x + 25 + x^{2} - 25 x}{5 x (x - 5)} = \frac{x^{2} - 10 x + 25}{5 x (x - 5)} = \frac{{(x - 5)}^{2}}{5 x (x - 5)} = \frac{x - 5}{5 x}$

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

e) $\frac{4 x^{2} - 3 x + 17}{x^{3} - 1} + \frac{2 x - 1}{x^{2} + x + 1} + \frac{6}{1 - x} = \frac{4 x^{2} - 3 x + 17}{(x - 1) (x^{2} + x + 1)} + \frac{2 x - 1}{x^{2} + x + 1} + \frac{- 6}{x - 1} = \frac{4 x^{2} - 3 x + 17}{(x - 1) (x^{2} + x + 1)} + \frac{(2 x - 1) (x - 1)}{(x - 1) (x^{2} + x + 1)} + \frac{- 6 (x^{2} + x + 1)}{(x - 1) (x^{2} + x + 1)} = \frac{4 x^{2} - 3 x + 17 + 2 x^{2} - 3 x + 1 - 6 x^{2} - 6 x - 6}{(x - 1) (x^{2} + x + 1)} = \frac{- 12 x + 12}{(x - 1) (x^{2} + x + 1)} = \frac{- 12 (x - 1)}{(x - 1) (x^{2} + x + 1)} = \frac{- 12}{x^{2} + x + 1}$