Bài 32 trang 50 SGK Toán 8 tập 1

Đố. Đố em tính nhanh được tổng sau:

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

Ta có:

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

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

..................

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

Do đó:

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

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

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

\(= \dfrac{{x + 6}}{{x\left( {x + 6} \right)}} + \dfrac{{ - x}}{{x\left( {x + 6} \right)}}\)

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