Phương trình tích
Phương trình tích có dạng $A\left( x \right).B\left( x \right)=0$
Công thức: $A(x).B(x)=0\Leftrightarrow A(x)=0$ hoặc $B\left( x \right)=0$
Tức là muốn giải phương trình $A\left( x \right).B\left( x \right)=0$, ta giải hai phương trình $A\left( x \right)=0$ và $B\left( x \right)=0$
Ví dụ: $\left( {x - 4} \right)\left( {x + 1} \right) = 0 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{x - 4 = 0}\\{x + 1 = 0}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{x = 4}\\{x = - 1}\end{array}} \right.$
$ (3,5\text{x}-7)(2,1\text{x}-6,3)=0\Leftrightarrow \left[ \begin{array}{l} 3,5\text{x}-7=0 \\ 2,1\text{x}-6,3=0 \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=2 \\ x=3 \end{array} \right.\Rightarrow S=2+3=5 $
$ \begin{array}{l} (x-2)(x+2)=0 \\ \Leftrightarrow \left[ \begin{array}{l} x-2=0 \\ x+2=0 \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=2. \\ x=-2. \end{array} \right. \end{array} $
$ \begin{array}{l} \left( 3-2\text{x} \right)\left( 6\text{x}+4 \right)\left( 5-8\text{x} \right)=0 \\ \Leftrightarrow \left[ \begin{array}{l} 3-2\text{x}=0 \\ 6\text{x}+4=0 \\ 5-8\text{x}=0 \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=\dfrac{3}{2} \\ x=-\dfrac{2}{3} \\ x=\dfrac{5}{8} \end{array} \right. \end{array} $
$ \begin{array}{l} \left( \sqrt{13}+5x \right)\left( 3,4-4x\sqrt{1,7} \right)=0 \\ \Leftrightarrow \left[ \begin{array}{l} \sqrt{13}+5x=0 \\ 3,4-4x\sqrt{1,7}=0 \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=-\dfrac{\sqrt{13}}{5}\approx -0,721 \\ x=\dfrac{3,4}{4\sqrt{1,7}}\approx 0,652 \end{array} \right. \end{array} $
$ (4x-10)(24+5x)=0\Leftrightarrow \left[ \begin{array}{l} 4x-10=0 \\ 24+5x=0 \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=\dfrac{5}{2} \\ x=-\dfrac{24}{5} \end{array} \right. $