Question

Tìm x, biết

\(\frac{x+1}{10}\)+ \(\frac{x+1}{11}\)+ \(\frac{x+1}{12}\)=\(\frac{x+1}{13}\)+\(\frac{x+1}{14}\)

 

\(\frac{x+4}{2000}\)+ \(\frac{x+3}{2001}\)= \(\frac{x+2}{2002}\)+ \(\frac{x+1}{2003}\)

Answer 1

\(\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)-\left(\frac{x+2}{2002}+1\right)-\left(\frac{x+1}{2003}+1\right)=0\)

\(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)

\(\left(x+2004\right).\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

Vì \(\frac{1}{2000}>\frac{1}{2001}>\frac{1}{2002}>\frac{1}{2003}\)

=>\(x+2004=0\)

=>\(x=-2004\)

Answer 2

a, Ta có \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{1+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

       =>  \(\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=\left(x+1\right).\left(\frac{1}{13}+\frac{1}{14}\right)\)

      =>   \(\left(x+1\right).\frac{181}{660}=\left(x+1\right).\frac{27}{182}\)

      =>    \(x.\frac{181}{660}+\frac{181}{660}=x.\frac{27}{182}+\frac{27}{182}\)

      =>   \(x\frac{181}{660}-x\frac{27}{182}=\frac{27}{182}-\frac{181}{660}\) 

      =>   \(x\frac{7561}{60060}=-\frac{7561}{60060}\)

     =>  x = -1

Mk chỉ biết cách làm này thôi mặc dù có cách khác nữa ( bạn thông càm ) 

Tích cho mk nhé