1, |2x-1|=5
⇒\(\left[ \begin{array}{l}2x-1=5\\2x-1=-5\end{array} \right.\)
⇒\(\left[ \begin{array}{l}2x=6\\2x=-4\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=3\\x=-2\end{array} \right.\)
2, $x^{2}$ = 9
⇒$x^{2}$ =±$3^{2}$
⇒ x = ±3
3, |3x+1|=x-2
⇒\(\left[ \begin{array}{l}3x+1=x-2\\3x+1=2-x\end{array} \right.\)
⇒\(\left[ \begin{array}{l}2x=-3\\4x=1\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=$\frac{-3}{2}$ \\x=$\frac{1}{4}$ \end{array} \right.\)
4, $x^{2}$ - 5=0
⇒ $x^{2}$ = 5
⇒ $x^{2}$ = ±$(\sqrt[]{5})^{2}$
⇒ x = ±$\sqrt[]{5}$
5, |2x-1|=5-x
⇒\(\left[ \begin{array}{l}2x-1=5-x\\2x-1=x-5\end{array} \right.\)
⇒\(\left[ \begin{array}{l}3x=6\\x=-4\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=2\\x=-4\end{array} \right.\)
6, $x^{2}$ + 4 = 0
⇒ $x^{2}$ = -4
Mà $x^{2}$ ≥ 0
⇒phương trình vô nghiệm
