a,
Điều kiện: `x+2>=0<=>x>= -2`
`|2x+3|=x+2`
$⇔\left[\begin{array}{l}2x+3=x+2\\2x+3=-x-2\end{array}\right.$
$⇔\left[\begin{array}{l}x=-1\\3x=-5\end{array}\right.$
$⇔\left[\begin{array}{l}x=-1(t/m)\\x=-\dfrac{5}{3}(t/m)\end{array}\right.$
Vậy `x∈{-1;-5/3}`
b,
$A=|x-2006|+|2007-x|\ge |x-2006+2007-x|=|1|=1$
Đẳng thức xảy ra $⇔(x-2006)(2007-x)\ge0$
$⇔(x-2006)(x-2007)\le0$
Vì $x-2006>x-2007$
$⇒\begin{cases}x-2006≥0\\x-2007≤0\end{cases}$
$⇔\begin{cases}x≥2006\\x≤2007\end{cases}$
$⇔2006≤x≤2007$
Vậy $A_{min}=1⇔2006≤x≤2007$
