Đáp án:
c, `|2x - 1| - |x + 1/3| = 0`
`|2x - 1| = |x + 1/3|`
⇒\(\left[ \begin{array}{l}2x -1 = x + \frac{1}{3}\\2x - 1 = -x - \frac{1}{3}\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x = \frac{4}{3}\\x = \frac{3}{5}\end{array} \right.\)
Vậy `x \in {4/3 ; 3/5}`
d, `3x - | x + 15| = 5/4`
` | x + 15| = 3x - 5/4`
⇒\(\left[ \begin{array}{l}x + 15 = 3x - \frac{5}{4}\\x + 15 = -3x + \frac{5}{4}\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x = -\frac{60}{19}\\x =- \frac{55}{16}\end{array} \right.\)
Vậy `x \in {-60/19 ; -55/16}`
