Question

\(\frac{1}{x\left(x+1\right)}=\frac{1}{x}+\frac{1}{2011}\)
Tính giá trị của \(x\)

Answer 1

Ta có:\(\frac{1}{x\left(x+1\right)}=\frac{1}{x}+\frac{1}{2011}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x}+\frac{1}{2011}\)

\(\Leftrightarrow-\frac{1}{x+1}=\frac{1}{2011}\)\(\Leftrightarrow-x-1=2011\)

\(\Leftrightarrow x=-2012\)

Answer 2

\(\frac{1}{x\left(x+1\right)}=\frac{1}{x}+\frac{1}{2011}\)

=> \(\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x}-\frac{-1}{2011}\)

=> \(\frac{1}{x+1}=\frac{-1}{2011}=\frac{1}{-2011}\)

=> x + 1 = -2011

=> x = -2011 - 1

=> x = -2012

Vậy x = -2012

Answer 3

Ta có: \(\frac{1}{x\left(x+1\right)}=\frac{1}{x}+\frac{1}{2011}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x}+\frac{1}{2011}\)

\(\Leftrightarrow\frac{1}{x+1}=-\frac{1}{2011}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{-2011}\)

\(\Leftrightarrow x+1=-2011\)

\(\Leftrightarrow x=-2012\)

Vậy \(x=-2012\)