Đáp án:
`a.`
`|2x - 1| = |x|`
`->` \(\left[ \begin{array}{l}2x-1=x\\2x-1=-x\end{array} \right.\)
`->` \(\left[ \begin{array}{l}2x-x=1\\2x+x=1\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=1\\3x=1\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=1\\x=\dfrac{1}{3}\end{array} \right.\)
Vậy `x=1` hoặc `x=1/3`
Vậy `x ∈ {1;1/3}`
`b,`
`|x-5| = |3x|`
`->` \(\left[ \begin{array}{l}x-5=3x\\x-5=-3x\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x-3x=5\\x+3x=5\end{array} \right.\)
`->` \(\left[ \begin{array}{l}-2x=5\\4x=5\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=\dfrac{-5}{2}\\x=\dfrac{5}{4}\end{array} \right.\)
Vậy `x ∈ {(-5)/2; 5/4}`
`c,`
`|5 - x| = |3x|`
`->` \(\left[ \begin{array}{l}5-x=3x\\5-x=-3x\end{array} \right.\)
`->` \(\left[ \begin{array}{l}5-x-3x=0\\5-x+3x=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}5-4x=0\\5 + 2x=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}4x=5\\2x=-5\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=\dfrac{5}{4}\\x=\dfrac{-5}{2}\end{array} \right.\)
Vậy `x ∈ {5/4; (-5)/2}`
`d,`
`|3x + 2| = |-x|`
`-> |3x + 2| = x` `(1)`
Điều kiện : $x \geqslant 0$
Thì `(1)` trở thành :
`|3x + 2|= |x|`
`->` \(\left[ \begin{array}{l}3x+2=x\\3x+2=-x\end{array} \right.\)
`->` \(\left[ \begin{array}{l}3x-x=-2\\3x+x=-2\end{array} \right.\)
`->` \(\left[ \begin{array}{l}2x=-2\\4x=-2\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=-1 \text{(Không thỏa mãn)}\\x=\dfrac{-1}{2} \text{(Không thỏa mãn)}\end{array} \right.\)
Vậy không có `x` thỏa mãn
