`a) |2x+3|=x+2`
`<=>`\(\left[ \begin{array}{l}2x+3=x+2\\2x+3=-x-2\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}2x-x=2-3\\2x+x=-2-3\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-1\\3x=-5\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-1\\x= \frac{-5}{3}\end{array} \right.\)
`b) A=|x-2006|+|2007-x|`
Áp dụng bđt `|a|+|b|>=|a+b|` ta có:
`A=|x-2006|+|2007-x|>=|x-2006+2007-x|=1`
dấu = có khi `(x-2006)(2007-x)>=0<=>2006<=x<=2007`
vậy `min A=1` khi `2006<=x<=2007`
