Giải bài 35 trang 50 – SGK Toán lớp 8 tập 1

 a) $\dfrac{x + 1}{x - 3} - \dfrac{1 - x}{x + 3} - \dfrac{2x(1 - x)}{9 - x^2}$

$= \dfrac{x + 1}{x - 3} - \dfrac{1 - x}{x + 3} - \dfrac{2x(1 - x)}{-(x^2 - 9)}$

$= \dfrac{x + 1}{x - 3} - \dfrac{1 - x}{x + 3} + \dfrac{2x(1 - x)}{x^2 - 9}$

$= \dfrac{(x + 1)(x + 3)}{(x - 3)(x + 3)} - \dfrac{(1 - x)(x - 3)}{(x + 3)(x - 3)} + \dfrac{2x(1 - x)}{(x - 3)(x + 3)}$

$= \dfrac{x^2 + 3x + x + 3}{(x - 3)(x + 3)} - \dfrac{x - 3 - x^2 + 3x}{(x + 3)(x - 3)} + \dfrac{2x - 2x^2}{(x - 3)(x + 3)}$

$= \dfrac{x^2 + 4x + 3}{(x - 3)(x + 3)} - \dfrac{-x^2 + 4x - 3}{(x + 3)(x - 3)} + \dfrac{2x - 2x^2}{(x - 3)(x + 3)}$

$= \dfrac{x^2 + 4x + 3 - (-x^2 + 4x - 3) + (2x - 2x^2)}{(x - 3)(x + 3)}$

$= \dfrac{x^2 + 4x + 3 +x^2 - 4x + 3+ 2x - 2x^2}{(x - 3)(x + 3)}$

$= \dfrac{ 2x + 6 }{(x - 3)(x + 3)}$

$= \dfrac{ 2(x + 3) }{(x - 3)(x + 3)}$

$= \dfrac{ 2 }{x - 3}$

b) $\dfrac{3x + 1}{(x - 1)^2} - \dfrac{1}{x + 1} + \dfrac{x + 3}{1 - x^2}$

$= \dfrac{3x + 1}{(x - 1)^2} - \dfrac{1}{x + 1} + \dfrac{x + 3}{-(x^2 - 1)}$

$= \dfrac{3x + 1}{(x - 1)^2} - \dfrac{1}{x + 1} - \dfrac{x + 3}{x^2 - 1}$

$= \dfrac{(3x + 1)(x + 1)}{(x - 1)^2(x + 1)} - \dfrac{(x - 1)^2}{(x - 1)^2(x + 1)} - \dfrac{(x + 3)(x - 1)}{(x - 1)(x + 1)(x - 1)}$

$= \dfrac{3x^2 + 3x + x + 1}{(x - 1)^2(x + 1)} - \dfrac{x^2 - 2x + 1}{(x - 1)^2(x + 1)} - \dfrac{x^2 - x + 3x - 3}{(x - 1)^2(x + 1)}$

$= \dfrac{3x^2 + 4x + 1}{(x - 1)^2(x + 1)} - \dfrac{x^2 - 2x + 1}{(x - 1)^2(x + 1)} - \dfrac{x^2 +2x - 3}{(x - 1)^2(x + 1)}$

$= \dfrac{3x^2 + 4x + 1 - (x^2 - 2x + 1) - (x^2 + 2x - 3)}{(x - 1)^2(x + 1)}$

$= \dfrac{3x^2 + 4x + 1 - x^2 + 2x - 1 - x^2 - 2x + 3}{(x - 1)^2(x + 1)}$

$= \dfrac{x^2 + 4x + 3}{(x - 1)^2(x + 1)}$

$= \dfrac{x^2 + x + 3x + 3}{(x - 1)^2(x + 1)}$

$= \dfrac{x(x + 1) + 3(x + 1)}{(x - 1)^2(x + 1)}$

$= \dfrac{(x + 1)(x + 3)}{(x - 1)^2(x + 1)}$

$= \dfrac{x + 3}{(x - 1)^2}$

Lưu ý: $\dfrac{A}{-B} = - \dfrac{A}{B}$, ví dụ $= \dfrac{1}{-(x - 3)} = - \dfrac{1}{x - 3}$