Question

Tính

\(A=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+...+\dfrac{1}{\left(x+9\right)\left(x+10\right)}\)

Answer 1

Đã ra vầy thì quá đơn giản :V, chẻ xuống

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

Tương tự rồi khử là ra kết quả ;V

Answer 2

A=\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+...+\dfrac{1}{\left(x+9\right)\left(x+10\right)}\)= \(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+9}-\dfrac{1}{x+10}\)

= \(\dfrac{1}{x}-\dfrac{1}{x+10}\)

= \(\dfrac{x+10}{x\left(x+10\right)}-\dfrac{x}{x\left(x+10\right)}\)

= \(\dfrac{10}{x\left(x+10\right)}\)