Question

Giá trị của x thỏa mãn: \(\frac{1}{x\left(x+1\right)}=\frac{1}{x}+\frac{1}{2011}\)

Ghi rõ cách giải ra

Answer 1

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

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

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

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

=>\(0-\frac{1}{x+1}=\frac{1}{2011}\)

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

=>-x+1=2011

=>-x=2011-1

=>-x=2010

=>x=-2010

Vậy x=-2010

Answer 2

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

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

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

<=>-x-1=2011

<=>x=-2012

Đáp số: \(x=-2012\)

Answer 3

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

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

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

<=>2011(-1)=x.1

<=>-2011=x

<=>x=-2011

vậy x=-2011

Answer 4

Sorry mí bạn, mk đọc nhầm nên lỡ k cho Nguyễn Mạnh Tuấn, Nguyễn Mạnh Tuấn giải sai ròi