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`a,`
`|x-1| = |2x-3|`
`->` \(\left[ \begin{array}{l}x-1=2x-3\\x-1=-2x+3\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}x-2x=1-3\\x+2x=1+3\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}-x=-2\\3x=4\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}x=2\\x=4:3\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}x=2\\x=\dfrac{4}{3}\end{array} \right.\)
Vậy `x=2` hoặc `x=4/3`
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`b,`
`|x+1| + |y-2|=0`
Với mọi `x,y` có : $\begin{cases} |x+1|≥0\\|y-2|≥0 \end{cases}$
`-> |x+1| + |y-2| ≥0∀x,y`
Dấu "`=`" xảy ra khi :
`↔` $\begin{cases} |x+1|=0\\|y-2|=0 \end{cases}$
`↔` $\begin{cases} x+1=0\\y-2=0 \end{cases}$
`↔` $\begin{cases} x=0-1\\y=0+2\end{cases}$
`↔` $\begin{cases} x=-1\\y=2\end{cases}$
Vậy `(x;y)=(-1;2)`
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`c,`
`|2x-1| -|x+3|=0`
`-> |2x-1|=0+|x+3|`
`-> |2x-1|=|x+3|`
`->` \(\left[ \begin{array}{l}2x-1=x+3\\2x-1=-x-3\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}2x-x=1+3\\2x+x=1-3\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}x=4\\3x=-2\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}x=4\\x=-2:3\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}x=4\\x=\dfrac{-2}{3}\end{array} \right.\)
Vậy `x=4` hoặc `x=(-2)/3`
