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`a,`
`|3x-9| = |x+2|`
`->` \(\left[ \begin{array}{l}3x-9=x+2\\3x-9=-x-2\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}3x-x=9+2\\3x+x=9-2\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}2x=11\\4x=7\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}x=\dfrac{11}{2}\\x=\dfrac{7}{4}\end{array} \right.\)
Vậy `x=11/2` hoặc `x=7/4`
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`b,`
`|5x-3| = |3x+2|`
`->` \(\left[ \begin{array}{l}5x-3=3x+2\\5x-3=-3x-2\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}5x-3x=3+2\\5x+3x=3-2\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}2x=5\\8x=1\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}x=\dfrac{5}{2}\\x=\dfrac{1}{8}\end{array} \right.\)
Vậy `x=5/2` hoặc `x=1/8`
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`c,`
`|3/2x +1/2| = |1/4x - 1|`
`->` \(\left[ \begin{array}{l}\dfrac{3}{2}x+\dfrac{1}{2}=\dfrac{1}{4}x-1\\\dfrac{3}{2}x+\dfrac{1}{2}=\dfrac{-1}{4}x+1\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}\dfrac{3}{2}x-\dfrac{1}{4}x=\dfrac{-1}{2}-1\\\dfrac{3}{2}x+\dfrac{1}{4}x=\dfrac{-1}{2}+1\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}\dfrac{5}{4}x=\dfrac{-3}{2}\\\dfrac{7}{4}x=\dfrac{1}{2}\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}x=\dfrac{-6}{5}\\x=\dfrac{2}{7}\end{array} \right.\)
Vậy `x=(-6)/5` hoặc `x=2/7`
