Đáp án:
Giải thích các bước giải:
$\text{1) 3x²-12x=0}$
$\text{⇔3x(x-4)=0}$
$\text{⇔\(\left[ \begin{array}{l}3x=0\\x-4=0\end{array} \right.\) }$
$\text{⇔\(\left[ \begin{array}{l}x=0\\x=4\end{array} \right.\) }$
$\text{Vậy S= {0; 4} }$
$\text{2) (x²-8x)-3x+24=0}$
$\text{⇔x(x-8)-3(x-8)=0}$
$\text{⇔(x-3)(x-8)=0}$
$\text{⇔\(\left[ \begin{array}{l}x-3=0\\x-8=0\end{array} \right.\) }$
$\text{⇔\(\left[ \begin{array}{l}x=3\\x=8\end{array} \right.\) }$
$\text{Vậy S= {3; 8} }$
$\text{3) (x-2)²-5(2-x)=0}$
$\text{⇔(x-2)²+5(x-2)=0}$
$\text{⇔(x-2)(x-2+5)=0}$
$\text{⇔(x-2)(x+3)=0}$
$\text{⇔\(\left[ \begin{array}{l}x-2=0\\x+3=0\end{array} \right.\) }$
$\text{⇔\(\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\) }$
$\text{Vậy S= {-3; 2} }$
$\text{4) (x³-8)+2x²-4x=0}$
$\text{⇔(x-2)(x²+2x+4)+2x(x-2)=0}$
$\text{⇔(x-2)(x²+2x+4+2x)=0}$
$\text{⇔(x-2)(x²+4x+4)=0}$
$\text{⇔(x-2)(x+2)²=0}$
$\text{⇔\(\left[ \begin{array}{l}x-2=0\\x+2=0\end{array} \right.\) }$
$\text{⇔\(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\) }$
$\text{Vậy S= {2; -2} }$
$\text{5) x²(x-3)+18-6x=0}$
$\text{⇔x²(x-3)-6(x-3)=0}$
$\text{⇔(x-3)(x²-6)=0}$
$\text{⇔\(\left[ \begin{array}{l}x-3=0\\x²-6=0\end{array} \right.\) }$
$\text{⇔\(\left[ \begin{array}{l}x=3\\x=±\sqrt{6}\end{array} \right.\) }$
$\text{Vậy S= {3; ±$\sqrt{6}$ } }$
