a, | 2x - 5 | = 3
⇒ \(\left[ \begin{array}{l}2x-5=3\\2x-5=-3\end{array} \right.\)
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b, 6 + |3x+1| =5x-12
*TH1: |3x+1|= 3x+1 ⇔ 3x+1≥0 ⇔ x ≥ $\frac{-1}{3}$
⇒ 6 + 3x+1 = 5x-12
⇔ -2x= -19
⇔ x= $\frac{19}{2}$ (t/m)
*TH2: |3x+1|= -3x-1 ⇔ 3x+1<0 ⇔ x < $\frac{-1}{3}$ \
⇒ 6 - 3x-1 = 5x-12
⇔ -8x = -17
⇔ x = $\frac{17}{8}$ (ko t/m)
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c) |1-5y|=5y-1
⇒ \(\left[ \begin{array}{l}1-5y=5y-1\\5y-1=5y-1\end{array} \right.\) ⇒ \(\left[ \begin{array}{l}y=-1/5\\y ∈ Z\end{array} \right.\)
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d, |2x-4|=|x ² -2x|
*TH1: 2x - 4 = x² - 2x
⇒ -x² + 4x - 4 = 0
⇒ -( x-2 )²=0
⇒ x-2=0
⇒ x=2
*TH2: 2x - 4 = -x² + 2x
⇒ x² - 4 = 0
⇒ ( x-2 ).( x+2 )=0
⇒\(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\)
Vậy...
