`a) |x-1|=2`
`<=>`\(\left[ \begin{array}{l}x-1=2\\x-1=-2\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=3\\x=-1\end{array} \right.\)
Vậy `S={3;-1}`
`b) |2x-4|+|x-1|=0`
Nếu `x>2` thì
`2x-4+x-1=0`
`<=> 3x=5`
`<=> x=5/3` (loại)
Nếu `1<=x<=2` thì
`4-2x+x-1=0`
`<=> -x+3=0`
`<=> x=3` (loại)
Nếu `x<1`
`4-2x+1-x=0`
`<=> 5-3x=0`
`<=> 3x=5`
`<=> x=5/3` (loại)
Vậy `S=∅`
`c) |3-x|=|2x-5|`
`<=>`\(\left[ \begin{array}{l}3-x=2x-5\\3-x=5-2x\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}2x+x=3+5\\-x+2x=5-3\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}3x=8\\x=2\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{8}{3}\\x=2\end{array} \right.\)
Vậy `S={8/3;2}`
`d) |x^2-1|=x-2`
$⇔\begin{cases}x-2≥0\\(x^2-1)^2=(x-2)^2\end{cases}$
$⇔\begin{cases}x≥2\\(x^2-1)^2-(x-2)^2=0\end{cases}$
$⇔\begin{cases}x≥2\\(x^2-1-x+2)(x^2-1+x-2)=0\end{cases}$
$⇔\begin{cases}x≥2\\(x^2-x+1)(x^2+x-3)=0\end{cases}$
$⇔\begin{cases}x≥2\\x^2+x-3=0\end{cases}$
$⇔\begin{cases}x≥2\\\left[ \begin{array}{l}x=\dfrac{-1+\sqrt{13}}{2}\\x=\dfrac{-1-\sqrt{13}}{2}\end{array} \right.\end{cases}$
`->` Pt vô nghiệm
Vậy `S=∅`
