Question

Giải PT

\(\frac{x+1}{2004}\)+ \(\frac{x+2}{2003}\)= \(\frac{x+3}{2002}\)+ \(\frac{x+4}{2001}\)

Nhớ giải chi tiết giúp mk nha. Thankssssssssssssssssssss

Answer 1

\(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\) 

\(\Leftrightarrow\frac{x+2005}{2004}+\frac{x+2005}{2004}-\frac{x+2005}{2003}-\frac{x+2005}{2003}=0\)

 \(\Leftrightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)

\(\Leftrightarrow x+2005=0\Leftrightarrow x=-2005\) 

Answer 2

=> (x+1)/2004+1+(x+2)/2003+1=(x+3)/2002+1+(x+4)/2001+1
=> (x+2005)/2004+(x+2005)/2003=(x+2005)/2002+(x+2005)/2001
=> (x+2005)(1/2004+1/2003-1/2002-1/2001)=0
=> x+2005=0
=> x=-2005

Answer 3

\(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\Leftrightarrow\frac{x+1+2004}{2004}+\frac{x+2+2003}{2003}-\frac{x+3+2003}{2003}-\frac{x+4+2001}{2001}=0\)

Answer 4

\(\Rightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}-\frac{x+2005}{2002}-\frac{x+2005}{2001}=0\Leftrightarrow x+2005=0\Rightarrow x=-2005\left(mình\right)làmtiếpnha\)

Answer 5

Bn nguyễn kim thương

Làm đúng r nhé

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