Đáp án:
Giải thích các bước giải:
a, 3x(x²-4)+5(x²-4)=0
⇒(3x+5)(x²-4)=0
⇔\(\left[ \begin{array}{l}3x+5=0\\x^2-4=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{-5}{3}\\x=+-2\end{array} \right.\)
b, $x^{4}$ +$x^{2}$ =0
⇔$x^{2}$ ($x^{2}$ +1)=0
⇔\(\left[ \begin{array}{l}x^2=0\\x^2=-1(loại)\end{array} \right.\)
⇔x=0
c, $x^{3}$ -4x=0
⇔x($x^{2}$ -4)=0
⇔\(\left[ \begin{array}{l}x=0\\x=-2\\x=2\end{array} \right.\)
d, $x^{2}$ =2x
⇔$x^{2}$ -2x=0
⇔x(x-2)=0
⇔\(\left[ \begin{array}{l}x=0\\x=2\end{array} \right.\)
e, $(x-3)^{2}$ =5(x-3)
⇔$(x-3)^{2}$ -5(x-3)=0
⇔(x-3)(x-3-5)=0
⇔(x-3)(x-8)=0
⇔\(\left[ \begin{array}{l}x-3=0\\x-8=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=3\\x=8\end{array} \right.\)
f, $x^{3}$ -15x=0
⇔x($x^{2}$ -15)=0
⇔\(\left[ \begin{array}{l}x=0\\x^2=15\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=\sqrt15\\x=-\sqrt15\end{array} \right.\)
