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

Đề bài

Thực hiện các phép tính:

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

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

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

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

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