Question

các bn giúp mk bài này nhé

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x.\left(x+1\right)}\) (x thuộc Z) tính

Answer 1

Đặt A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x.\left(x+1\right)}\)

A=\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\)

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

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

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

Answer 2

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x\left(x+1\right)}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\)

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

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