Câu hỏi của Monkey D .Luffy

Question

Tìm x biết : $\dfrac{x+1}{2004}+\dfrac{x+2}{2003}+\dfrac{x+3}{2002}+35=2^5$

Lightning Farron · Accepted Answer

$\dfrac{x+1}{2004}+\dfrac{x+2}{2003}+\dfrac{x+3}{2002}+35=2^5$

$pt\Leftrightarrow\dfrac{x+1}{2004}+\dfrac{x+2}{2003}+\dfrac{x+3}{2002}+3=0$

$\Leftrightarrow\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1+\dfrac{x+3}{2002}+1=0$

$\Leftrightarrow\dfrac{x+1}{2004}+\dfrac{2004}{2004}+\dfrac{x+2}{2003}+\dfrac{2003}{2003}+\dfrac{x+3}{2002}+\dfrac{2002}{2002}=0$

$\Leftrightarrow\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}+\dfrac{x+2005}{2002}=0$

$\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}+\dfrac{1}{2002}\right)=0$

$\Rightarrow x+2005=0$. Do $\dfrac{1}{2004}+\dfrac{1}{2003}+\dfrac{1}{2002}\ne0$

$\Rightarrow x=-2005$Câu trả lời: $\dfrac{x+1}{2004}+\dfrac{x+2}{2003}+\dfrac{x+3}{2002}+35=2^5$