`|2-3x|+|2x-3y+1|=0`
Vì $\begin{cases} |2-3x|≥0∀x\\|2x-3y+1|≥0∀x,y \end{cases}$
`⇒|2-3x|+|2x-3y+1|≥0∀x,y`
Dấu "=" xảy ra $⇔\begin{cases} |2-3x|=0\\|2x-3y+1|=0 \end{cases}$
$⇔\begin{cases} 2-3x=0\\2x-3y+1=0 \end{cases}$
$⇔\begin{cases} x=\dfrac{2}{3}\\2.\dfrac{2}{3}-3y+1=0 \end{cases}$
$⇔\begin{cases} x=\dfrac{2}{3}\\\dfrac{4}{3}-3y+1=0 \end{cases}$
$⇔\begin{cases} x=\dfrac{2}{3}\\y=\dfrac{7}{9} \end{cases}$
