\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Rightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+3}{2001}+1\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
có 1/2000 + 1/2001 - 1/2002 - 1/2003
=> x + 2004 = 0
=> x = -2004
\(\frac{x+4}{2000}+\frac{x+3}{2001}+\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Leftrightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2004=0\)
\(\Leftrightarrow x=-2004\)
Ta có:
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
<=> \(\frac{x+4}{2000}+\frac{x+3}{2001}-\frac{x+2}{2002}-\frac{x+1}{2003}=0\)
<=> \(\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)-\left(\frac{x+2}{2002}+1\right)-\left(\frac{x+1}{2003}+1\right)=0\)
<=> \(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
<=> \(\left(x+2004\right).\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
<=> x+2004 =0 ( do \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)
<=> x= -2004
Học tốt
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Leftrightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\)\(\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}\)\(=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}\)\(-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right).\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2004=0\Leftrightarrow x=-2004\)
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Leftrightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\Leftrightarrow\left(\frac{x+2004}{2000}\right)+\left(\frac{x+2004}{2001}\right)=\left(\frac{x+2004}{2002}\right)+\left(\frac{x+2004}{2003}\right)\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2004=0\)
\(\Leftrightarrow x=-2004\)
Trl:
\(KQ=0\)
Hc tốt
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