Question

[ĐẶNG DUY KHƯƠNG]

1/x×{x+1)+1/(x+1)×(x+2)+1/(x+2)×(x+3)-1/x=1/2010

Answer 1

[Trần Thanh Phương]

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

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

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

\(\Rightarrow-\left(x-3\right)=2010\)

\(\Rightarrow x=-2013\)

Answer 2

[Nguyễn Phương Uyên]

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

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

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

\(\Rightarrow\left(\frac{1}{x}-\frac{1}{x}\right)-\frac{1}{x+3}=\frac{1}{2010}\)

\(\Rightarrow\frac{1}{x+3}=\frac{1}{2010}\)

\(\Rightarrow x=2007\)