a,
C = |x-2| + |5-x|≥ | x-2+5-x|= | -3|= 3
Dấu = xảy ra khi ( x-2).( 5-x)≥ 0
⇔ \(\left[ \begin{array}{l}\left \{ {{x-2≥0} \atop {5-x≥0}} \right.\\\left \{ {{x-2<0} \atop {5-x<0}} \right.\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}\left \{ {{x≥2} \atop {x≤5}} \right.\\\left \{ {{x<2} \atop {x>5}} \right.\end{array} \right.\)
Vậy Cmin= 3 khi 2≤x≤5
b, | 3x + 1 | > 4
⇔ \(\left[ \begin{array}{l}3x+1>4\\3x+1< -4\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x>1\\x<\frac{-5}{4}\end{array} \right.\)
c, | 5x - 3 | ≥ 7
⇔ \(\left[ \begin{array}{l}5x-3≥7\\5x-3≤-7\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x≥10\\x≤\frac{-4}{5}\end{array} \right.\)
