Đáp án + Giải thích các bước giải:
`1)`
`|12-x|=3`
`to [(12-x=3),(12-x=-3):} to [(x=9),(x=15):}`
Vậy `x in {9;15}`
`2)`
`|3x+4|=5`
`to [(3x+4=5),(3x+4=-5):} to [(3x=1),(3x=-9):}`
`to` \(\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=-3\end{array} \right.\)
Vậy `x in {-3;1/3}`
`3)`
`1/2+|x-2|=1`
`to |x-2|=1-1/2`
`to |x-2|=1/2`
`to` \(\left[ \begin{array}{l}x-2=\dfrac{1}{2}\\x-2=-\dfrac{1}{2}\end{array} \right.\) `to` \(\left[ \begin{array}{l}x=\dfrac{5}{2}\\x=\dfrac{3}{2}\end{array} \right.\)
Vậy `x in {3/2;5/2}`
`4)`
`|1/2x+1|-3/2=9`
`to |1/2x+1|=9+3/2`
`to |1/2x+1|=21/2`
`to` \(\left[ \begin{array}{l}\dfrac{1}{2}x+1=\dfrac{21}{2}\\\dfrac{1}{2}x+1=-\dfrac{21}{2}\end{array} \right.\) `to` \(\left[ \begin{array}{l}x=19\\x=-23\end{array} \right.\)
Vậy `x in {-23;19}`
`5)`
`2|3x-2|=8`
`to |3x-2|=4`
`to [(3x-2=4),(3x-2=-4):} to [(3x=6),(3x=-2):}`
`to` \(\left[ \begin{array}{l}x=2\\x=-\dfrac{2}{3}\end{array} \right.\)
Vậy `x in {-2/3;2}`
