Đáp án:
`a,`
`1,5 - |5/4 - 3x| = 3/4`
`-> |5/4 - 3x| = 1,5 - 3/4`
`-> |5/4 -3x| = 3/4`
`->` \(\left[ \begin{array}{l}\dfrac{5}{4}-3x=\dfrac{3}{4}\\ \dfrac{5}{4}-3x=\dfrac{-3}{4}\end{array} \right.\)
`->` \(\left[ \begin{array}{l}3x=\dfrac{1}{2}\\3x=2\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=\dfrac{1}{6}\\x=\dfrac{2}{3}\end{array} \right.\)
Vậy `x ∈ {1/6;2/3}`
`b,`
`|x + 1/2|=3`
`->` \(\left[ \begin{array}{l}x+\dfrac{1}{2}=3\\x+\dfrac{1}{2}=-3\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=\dfrac{5}{2}\\x=\dfrac{-7}{2}\end{array} \right.\)
Vậy `x ∈ {5/2; (-7)/2}`
`c,`
`|2x - 1| - |3x - 1| = 0`
`-> |2x - 1| = |3x - 1|`
`->` \(\left[ \begin{array}{l}2x-1=3x-1\\2x-1=-3x+1\end{array} \right.\)
`->` \(\left[ \begin{array}{l}-x=0\\5x=2\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=0\\x=\dfrac{2}{5}\end{array} \right.\)
Vậy `x ∈ {0;2/5}`
`d,`
`|3x - 1| - 2x = 1/2`
`-> |2x - 1|= 1/2 + 2x` `(1)`
Điều kiện : `1/2 + 2x` $\geqslant 0$ `-> 2x` $\geqslant$ `-1/2` `-> x` $\geqslant$ `-1/4`
Từ `(1)` ta có :
`-> |2x - 1| = |1/2 + 2x|`
`->` \(\left[ \begin{array}{l}2x-1=\dfrac{1}{2}+2x\\2x-1=-\dfrac{1}{2}-2x\end{array} \right.\)
`->` \(\left[ \begin{array}{l}0=\dfrac{3}{2} \text{(Vô lí)}\\x=\dfrac{1}{8} \text{(Thỏa mãn điềukiện)}\end{array} \right.\)
Vậy `x = 1/8`
`e,`
`|x - 1| - x = 1/2`
`-> |x - 1| = 1/2 +x` `(1)`
Điều kiện : `1/2 + x` $\geqslant 0$ `→ x` $\geqslant$ `-1/2`
Từ `(1)` trở thành :
`-> |x - 1| = |1/2 +x|`
`->` \(\left[ \begin{array}{l}x-1=\dfrac{1}{2}+x\\x-1=-\dfrac{1}{2}-x\end{array} \right.\)
`->` \(\left[ \begin{array}{l}0=\dfrac{1}{2} \text{(Vô lí)}\\x=\dfrac{1}{4} \text{(Thỏa mãn điều kiện)}\end{array} \right.\)
Vậy `x = 1/4`
