Giải bài 32 trang 50 - Sách giáo khoa Toán 8 tập 1

Đề bài

Đố. Đố em tính nhanh được tổng sau:

Hướng dẫn giải

Ta có : \(\dfrac{1}{x(x+1)}=\dfrac{(x+1)-x }{x(x+1)} = \dfrac{(x+1) }{x(x+1)} - \dfrac{x }{x(x+1)}= \dfrac{1}{x}-\dfrac{1}{x+1}\);

Tương tự : 

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

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

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

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

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

Do đó : 

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

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