[Toán 8]GPT: $\dfrac{5-x}{2003}-2=\dfrac{2-x}{2004}+\dfrac{5x}{2003}$

E

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[tex]\frac{{5 - x}}{{2003}} - 2 = \frac{{2 - x}}{{2004}} + \frac{{5x}}{{2003}}[/tex]
[tex] \Leftrightarrow \frac{5}{{2003}} - \frac{x}{{2003}} - 2 = \frac{2}{{2004}} - \frac{x}{{2004}} + 5.\frac{x}{{2003}}[/tex]
[tex] \Leftrightarrow \frac{x}{{2004}} - \frac{x}{{2003}} - 5.\frac{x}{{2003}} = \frac{2}{{2004}} + 2 - \frac{5}{{2003}}[/tex]
[tex] \Leftrightarrow x\left( {\frac{1}{{2004}} - \frac{1}{{2003}} - \frac{5}{{2003}}} \right) = \frac{1}{{1002}} + 2 - \frac{5}{{2003}}[/tex]
[tex] \Leftrightarrow x.\frac{{ - 10021}}{{4014012}} = \frac{{4011005}}{{2007006}}[/tex]
[tex] \Leftrightarrow x = \frac{{4011005}}{{2007006}}:\frac{{ - 10021}}{{4014012}}[/tex]
Số có vẻ to quá! Ai có cách nào hay hơn thì post lên nhé!:):)
 
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