Đáp án:
`a, S={-4;2/3}`
`b, S={-4;1}`
Giải thích các bước giải:
`a, |2x+1|=|x-3|`
`<=>`\(\left[ \begin{array}{l}2x+1=x-3\\2x+1=-(x-3)\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}2x-x=-3-1\\2x+x=3-1\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-4\\3x=2\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-4\\x=\dfrac{2}{3}\end{array} \right.\)
Vậy `S={-4;2/3}`
`b, |x²+2x+3|=7-x`
Vì `x²+2x+3≥0∀x∈R`
`=> x²+2x+3=7-x`
`<=> x²+3x-4=0`
`<=> (x-1)(x+4)=0`
`<=>`\(\left[ \begin{array}{l}x=1\\x=-4\end{array} \right.\) (TM)
Vậy `S={-4;1}`
