~ Bạn tham khảo ~
`a, |x+3|=|1-3x|`
`<=> (x+3)^2 = (1-3x)^2`
`<=> x^2+6x+9 = 1 - 6x + 9x^2`
`<=> x^2 - 9x^2 + 6x + 6x +9 - 1= 0`
`<=> -8x^2+12x+8=0`
`<=> -4(2x^2 - 3x - 2)=0`
`<=> -4(2x^2 + x - 4x - 2)=0`
`<=> -4[x(2x+1) - 2(2x+1)]=0`
`<=> -4(2x+1)(x-2)=0`
\(⇔\left[ \begin{array}{l}2x+1=0\\x-2=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=-\dfrac12\\x=2\end{array} \right.\)
Vậy `S={-1/2;2}`
`b,` Vì `|x+1|>=0;|x+2|>=0`
`=> |x+1|+|x+2|>=0`
`=> 3x>=0`
`=> x>=0`
Thay vào PT ta có :
`PT <=> x+1+x+2=3x`
`<=> x+x-3x=-1-2`
`<=> -x=-3`
`<=> x=3` (TMĐK)
Vậy `S={3}`
