`5)`
`A=|2x-1/3|-1 3/4`
`=|2x-1/3|-7/4`
Vì `|2x-1/3|>=0∀x`
`=>|2x-1/3|-7/4>= -7/4∀x`
`=>A>= -7/4`
Vậy `A_min=-7/4<=>|2x-1/3|=0<=>x=1/6`
`b)`
Vì $\begin{cases} \dfrac{1}{3}|x-2|>=0∀x\\2|3-\dfrac{1}{2}y|>=0∀y \end{cases}$
`=>1/3|x-2|+2|3-1/2y|+4>=4∀x;y`
`=>B>=4`
`=>B_min=4<=>`$\begin{cases} \dfrac{1}{3}|x-2|=0\\2|3-\dfrac{1}{2}y|=0 \end{cases}⇔\begin{cases} x=2\\y=6 \end{cases}$
