Question

giải phương trình: x/2004+(x+1)/2005+(x+2)/2006+(x+3)/2007=4

Answer 1

giúp mình nhanh vs ạ. mình cảm ơn

Answer 2

\(\Leftrightarrow\left(\frac{x}{2004}-1\right)+\left(\frac{x+1}{2005}-1\right)+\left(\frac{x+2}{2006}-1\right)+\left(\frac{x+3}{2007}-1\right)=0\)

\(\Leftrightarrow\frac{x-2004}{2004}+\frac{x-2004}{2005}+\frac{x-2004}{2006}+\frac{x-2004}{2007}=0\Leftrightarrow\left(x-2004\right)\left(...\right)=0\Rightarrow x=2004\)

Answer 3

\(\frac{x}{2004}+\frac{x+1}{2005}+\frac{x+2}{2006}+\frac{x+3}{2007}=4\)

⇔\(\frac{x}{2004}-1+\frac{x+1}{2005}-1+\frac{x+2}{2006}-1+\frac{x+3}{2007}-1=0\)

⇔\(\frac{x-2004}{2004}+\frac{x-2004}{2005}+\frac{x-2004}{2006}+\frac{x-2004}{2007}=0\)

⇔\(\left(x-2004\right)\left(\frac{1}{2004}+\frac{1}{2005}+\frac{1}{2006}+\frac{1}{2007}\right)=0\)

⇔\(x=2004\)

Chúc bạn học tốt!!!