Question

(x-2012)/8+(x-2008)/6+(x-2005)/5=10-(x-2004)/4

Answer 1

Ta có : \(\dfrac{x-2012}{8}+\dfrac{x-2008}{6}+\dfrac{x-2005}{5}=10-\dfrac{x-2004}{4}\)

\(\Leftrightarrow\left(\dfrac{x-2012}{8}-1\right)+\left(\dfrac{x-2008}{6}-2\right)+\left(\dfrac{x-2005}{5}-3\right)+\left(\dfrac{x-2004}{4}-4\right)=0\)\(\Leftrightarrow\dfrac{x-2020}{8}+\dfrac{x-2020}{6}+\dfrac{x-2020}{5}+\dfrac{x-2020}{4}=0\)

\(\Leftrightarrow\left(x-2020\right).\left(\dfrac{1}{8}+\dfrac{1}{6}+\dfrac{1}{5}+\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow x-2020=0\Leftrightarrow x=2020\)

Vậy x = 2020