Đáp án:
$\begin{array}{l}\textbf{Bài 1:}\\1. \ x^3-2x=0\\\leftrightarrow x(x^2-2)=0\\\leftrightarrow \left[\begin{array}{l}x=0\\x^2-2=0\end{array}\right.\\\leftrightarrow \left[\begin{array}{l}x=0\\x=\pm2\end{array}\right.\\2. \ x(x-4)+(x-4)=0\\\leftrightarrow (x-4)(x+1)=0\\\leftrightarrow \left[\begin{array}{l}x-4=0\\x+1=0\end{array}\right.\\\leftrightarrow \left[\begin{array}{l}x=4\\x=-1\end{array}\right.\\3. \ x(x-3)+4x-12=0\\\leftrightarrow x(x-3)+4(x-3)=0\\\leftrightarrow (x-3)(x+4)=0\\\leftrightarrow \left[\begin{array}{l}x-3=0\\x+4=0\end{array}\right.\\\leftrightarrow \left[\begin{array}{l}x=3\\x=-4\end{array}\right.\\4. \ 5x(x-2000)-x+2000=0\\\leftrightarrow (x-2000)(5x-1)=0\\\leftrightarrow \left[\begin{array}{l}x=2000\\x=\dfrac{1}{5}\end{array}\right.\\5. \ x^3-13x=0\\\leftrightarrow x(x^2-13)=0\\\leftrightarrow \left[\begin{array}{l}x=0\\x^2-13=0\end{array}\right.\\\leftrightarrow \left[\begin{array}{l}x=0\\x=\pm13\end{array}\right.\\\textbf{Bài 2:}\\b. \ x(x-1)-y(1-x)\\=x(x-1)+y(x-1)\\=(x-1)(x+y)\\\text{Thay} \ x=2001, y=1999 \ \text{vào biểu thức, ta được:}\\(2001-1)(2001+1999)\\=2000\times4000\\=8000000\end{array}$
