Giải bài 32 trang 50 – SGK Toán lớp 8 tập 1

Áp dụng kết quả bài 31 trang 50 sgk toán 8 tập 1 ta có:

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

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

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

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

$\dfrac{1}{(x + 4)(x + 5)} = \dfrac{1}{x + 4} - \dfrac{1}{x + 5}$

$\dfrac{1}{(x + 5)(x + 6)} = \dfrac{1}{x + 5} - \dfrac{1}{x + 6}$

Vậy

$\dfrac{1}{x(x + 1)} + \dfrac{1}{(x + 1)(x + 2)} + \dfrac{1}{(x + 2)(x + 3)} + \dfrac{1}{(x + 3)(x + 4)} + \dfrac{1}{(x + 4)(x + 5)} + \dfrac{1}{(x + 5)(x + 6)}$

$= \dfrac{1}{x} - \dfrac{1}{x + 1} + \dfrac{1}{x + 1} - \dfrac{1}{x + 2} + \dfrac{1}{x + 2} - \dfrac{1}{x + 3} + \dfrac{1}{x + 3} - \dfrac{1}{x + 4} + \dfrac{1}{x + 4} - \dfrac{1}{x + 5} + \dfrac{1}{x + 5} - \dfrac{1}{x + 6}$

$= \dfrac{1}{x} - \dfrac{1}{x + 6}$

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

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

$= \dfrac{6}{x(x + 6)}$

Lưu ý:  $\dfrac{1}{a(a + 1)} = \dfrac{1}{a} - \dfrac{1}{a + 1} \,\, \text{với}\,\, a \ne 0; \, a \ne -1$