*Bạn tham khảo nhé!!!!
a. (x + 2)(x - 3) = 0
⇔\(\left[ \begin{array}{l}x+2=0\\x-3=0\end{array} \right.\) ⇒ \(\left[ \begin{array}{l}x=-2\\x=3\end{array} \right.\)
b. (x - 5)(7 - x) = 0
⇔\(\left[ \begin{array}{l}x-5=0\\7-x=0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=5\\x=7\end{array} \right.\)
c. (x - 1)(x + 5)(-3x + 8) = 0
⇔\(\left[ \begin{array}{l}x-1=0\\x+5=0\\-3x+8=0\end{array} \right.\) ⇒ \(\left[ \begin{array}{l}x=1\\x=-5\\x=\frac{8}{3}\end{array} \right.\)
d. (x - 1)(3x + 1) = 0
⇔\(\left[ \begin{array}{l}x-1=0\\3x+1=0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=1\\x=\frac{-1}{3}\end{array} \right.\)
e. (x - 1)(x + 2)(x - 3) = 0
⇔\(\left[ \begin{array}{l}x-1=0\\x+2=0\\x-3=0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=1\\x=-2\\x=3\end{array} \right.\)
f. x(x² - 1) = 0
⇔x(x + 1)(x - 1) = 0
⇔\(\left[ \begin{array}{l}x=0\\x+1=0\\x-1=0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=0\\x=-1\\x=1\end{array} \right.\)
