# Giải bài 35 trang 50 - Sách giáo khoa Toán 8 tập 1

##### Hướng dẫn giải

a) Ta có $$9 - x^2 = -(x^2-9) = - (x-3)(x+3)$$

MTC = $$(x-3)(x+3)$$

Do đó : $$\dfrac{x + 1}{x – 3} – \dfrac{1 – x}{x + 3} – \dfrac{2x(1 – x)}{9 – x^2}$$

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

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

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

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

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

b) Ta có : $$1 - x^2 = (1-x)(1+x)$$

MTC = $$(1-x)^2(x+1)$$

Do đó : $$\dfrac{3x + 1}{(x – 1)^2} – \dfrac{1}{x + 1} + \dfrac{x + 3}{1 – x^2}$$

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

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

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

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