Đáp án:
a ) \(\left[ \begin{array}{l}x=0\\x=±3\end{array} \right.\)
b ) \(\left[ \begin{array}{l}x=1\\x=-\frac{5}{3}\end{array} \right.\)
Giải thích các bước giải:
a ) $x^{3}$ - 9x = 0
→ x( $x^{2}$ - 9 ) = 0
→ \(\left[ \begin{array}{l}x=0\\x^2-9=0\end{array} \right.\)
→ \(\left[ \begin{array}{l}x=0\\x^2=9\end{array} \right.\)
→ \(\left[ \begin{array}{l}x=0\\x=±3\end{array} \right.\)
b ) $(3x+1)^{2}$ - 16 = 0
→ $(3x+1)^{2}$ = 16
→ \(\left[ \begin{array}{l}(3x+1)^2=4^2\\(3x+1)^2=(-4)^2\end{array} \right.\)
→ \(\left[ \begin{array}{l}3x+1=4\\3x+1=-4\end{array} \right.\)
→ \(\left[ \begin{array}{l}3x=3\\3x=-5\end{array} \right.\)
→ \(\left[ \begin{array}{l}x=1\\x=-\frac{5}{3}\end{array} \right.\)
