Giải thích các bước giải:
c.Ta có:
$\dfrac{x+1}{19}+\dfrac{x+3}{17}+\dfrac{x+6}{14}=-3$
$\to (\dfrac{x+1}{19}+1)+(\dfrac{x+3}{17}+1)+(\dfrac{x+6}{14}+1)=0$
$\to \dfrac{x+1+19}{19}+\dfrac{x+3+17}{17}+\dfrac{x+6+14}{14}=0$
$\to \dfrac{x+20}{19}+\dfrac{x+20}{17}+\dfrac{x+20}{14}=0$
$\to (x+20)(\dfrac1{19}+\dfrac1{17}+\dfrac1{14})=0$
$\to x+20=0$
$\to x=-20$
d.Ta có:
$\dfrac{x-1}{19}+\dfrac{x-3}{17}+\dfrac{x-6}{14}=3$
$\to (\dfrac{x-1}{19}-1)+(\dfrac{x-3}{17}-1)+(\dfrac{x-6}{14}-1)=0$
$\to \dfrac{x-1-19}{19}+\dfrac{x-3-17}{17}+\dfrac{x-6-14}{14}=0$
$\to \dfrac{x-20}{19}+\dfrac{x-20}{17}+\dfrac{x-20}{14}=0$
$\to (x-20)(\dfrac1{19}+\dfrac1{17}+\dfrac1{14})=0$
$\to x=20$
