Giải bất phương trình sau: x+1/199 + x+2/198 + x+3/197 > -3

Ta có :

$\dfrac{x+1}{199}+\dfrac{x+2}{198}+\dfrac{x+3}{197} > -3$

$\to \bigg(\dfrac{x+1}{199}+1\bigg) +  \bigg(\dfrac{x+2}{198}+1\bigg) + \bigg(\dfrac{x+3}{197}+1\bigg)  >  0$

$\to \dfrac{x+200}{199}+\dfrac{x+200}{198}+\dfrac{x+200}{197} > 0 $

$\to (x+200).\bigg(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197} \bigg) > 0 $

Vì $\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197} > 0 $

Nên $x+200 > 0 $

$\to x> -200$

Vậy $x>-200$