Question

Giải bất phương trình:

\(\dfrac{x+4}{2016}+\dfrac{x+2}{2018}\ge\dfrac{x+14}{2006}+\dfrac{x+83}{1937}\)

Answer 1

\(\dfrac{x+4}{2016}+\dfrac{x+2}{2018}\ge\dfrac{x+14}{2006}+\dfrac{x+83}{1937}\)

\(\Leftrightarrow\dfrac{x+4}{2016}+1+\dfrac{x+2}{2018}+1\ge\dfrac{x+14}{2006}+1+\dfrac{x+83}{1937}+1\)

\(\Leftrightarrow\dfrac{x+2020}{2016}+\dfrac{x+2020}{2018}-\dfrac{x+2020}{2006}-\dfrac{x+2020}{1937}\ge0\)

\(\Leftrightarrow\left(x+2020\right)\left(\dfrac{1}{2016}+\dfrac{1}{2018}-\dfrac{1}{2006}-\dfrac{1}{1937}\right)\ge0\)

\(\Leftrightarrow x+2020\ge0\Leftrightarrow x\ge-2020\)

Vậy \(x\ge-2020\)