Question

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

 

Answer 1

Hồ Thu Giang có nhầm lẫn rồi , sai bắt đầu từ dòng thứ 3:

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

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

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

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

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

\(\Rightarrow x=-2012\)

Answer 2

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

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

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

=> \(\frac{1}{x+1}=\frac{1}{2011}\)

=> x+1 = 2011

=> x = 2011 - 1

=> x = 2010

Answer 3

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

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

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

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

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

\(\Rightarrow x=-2012\)

Answer 4

bạn ơi cho mình hỏi tại sao từ 1 / x(x+1) => 1/x-1/x+1 ?