Question

Giải phương trình:

 \(\frac{x+1}{2005}+\frac{x+1}{2003}=\frac{x+1}{2001}+\frac{x+1}{1999}\)

Answer 1

\(pt\Leftrightarrow\left(x+1\right)\left(\frac{1}{2005}+\frac{1}{2003}-\frac{1}{2001}-\frac{1}{1999}\right)=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Answer 2

\(\frac{x+1}{2005}+\frac{x+1}{2003}=\frac{x+1}{2001}+\frac{x+1}{1999}.\)

\(\Rightarrow\frac{x+1}{2005}+\frac{x+1}{2003}-\frac{x+1}{2001}-\frac{x+1}{1999}=0\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{2005}+\frac{1}{2003}-\frac{1}{2001}-\frac{1}{1999}\right)=0\)

Mà \(\frac{1}{2005}+\frac{1}{2003}-\frac{1}{2001}-\frac{1}{1999}#0\)

\(\Rightarrow x+1=0\Rightarrow x=-1\)

Vậy nghiệm của pt là x = -1

Answer 3

\(\frac{x+1}{2005}+\frac{x+1}{2003}=\frac{x+1}{2001}+\frac{x+1}{1999}\)

\(\Rightarrow\frac{x+1}{2005}+\frac{x+1}{2003}-\frac{x+1}{2001}-\frac{x+1}{1999}=0\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{2005}+\frac{1}{2003}-\frac{1}{2001}-\frac{1}{1999}\right)=0\)

Do : \(\frac{1}{2015}+\frac{1}{2003}-\frac{1}{2001}-\frac{1}{1999}\)không thể bằng 0

\(\Rightarrow x+1=0\)

\(\Rightarrow x=0-1=-1\)