Đáp án:
a, Ta có :
$| 4x + 1| = 1/2$
<=> \(\left[ \begin{array}{l}4x + 1 = 1/2\\4x + 1 = -1/2\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x = -1/8\\x=-3/8\end{array} \right.\)
b, Ta có :
$|4x + 2| = 0$
$ <=> 4x + 2 = 0$
$ <=> x = -1/2$
c, Ta có :
$|1/2x - 3| = 2/3$
<=> \(\left[ \begin{array}{l}1/2x - 3 = 2/3\\1/2x - 3 = -2/3\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=22/3\\x=14/3\end{array} \right.\)
d, Ta có :
$( x - 3)^2 = 1/4$
<=> \(\left[ \begin{array}{l}x - 3 = 1/2\\x - 3 = -1/2\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=7/2\\x=5/2\end{array} \right.\)
Giải thích các bước giải:
