Dạng 2. ...
a. | x | = | -5 |
⇔ | x | = 5
⇒ \(\left[ \begin{array}{l}x=5.\\x=-5.\end{array} \right.\)
Vậy...
----
b. | x - 1 | = 4
⇔ \(\left[ \begin{array}{l}x-1=4\\x-1=-4\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=5.\\x=-3.\end{array} \right.\)
Vậy...
----
c. | x + 3 | = 0
⇔ x + 3 = 0
⇔ x = 0 - 3
⇔ x = -3.
Vậy...
----
d. | x + 2 | = 12 + (-3) + | -4 |
⇔ | x + 2 | = 9 + 4
⇔ | x + 2 | = 13
⇔ \(\left[ \begin{array}{l}x+2=13\\x+2=-13\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=11.\\x=-15.\end{array} \right.\)
Vậy...
