Question

Tìm x biết :\(\dfrac{1}{x\left(x+1\right)}\) +\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}\) \(\dfrac{1}{\left(x+2\right)\left(x+3\right)}\) -\(\dfrac{1}{x}\) =\(\dfrac{1}{2010}\)

Answer 1

\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}-\dfrac{1}{x}=2010\)

\(\Rightarrow\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}-\dfrac{1}{x}=\dfrac{1}{2010}\)\(\Rightarrow\dfrac{-1}{x+3}=\dfrac{1}{2010}\)

\(\Rightarrow x+3=-2010\)

\(\Rightarrow x=-2013\)

Vậy x = -2013

Answer 2

\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}-\dfrac{1}{x}=\dfrac{1}{2010}\)

=> \(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}-\dfrac{1}{x}=\dfrac{1}{2010}\)

=> \(-\dfrac{1}{x+3}=\dfrac{1}{2010}\)

=> \(-2010=x+3\)

=> \(x=-2013\)