Question

Answer 1

\(5,\left(3x-6\right)\left(7-10x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-6=0\\7-10x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{10}\end{matrix}\right.\)

\(6,\left(6x+7\right)\left(3x-2\right)\left(2x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}6x+7=0\\3x-2=0\\2x-9=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{6}\\x=\dfrac{2}{3}\\x=\dfrac{9}{2}\end{matrix}\right.\)

\(7,\left(3,4x-6,8\right)\left(15x+2,5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3,4x-6,8=0\\15x+2,5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{6}\end{matrix}\right.\)

Answer 2

\(5)\left(3x-6\right)\left(7-10x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-6=0\\7-10x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=6\\10x=7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{10}\end{matrix}\right.\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{2;\dfrac{7}{10}\right\}\)

\(6)\left(6x+7\right)\left(3x-2\right)\left(2x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}6x+7=0\\3x-2=0\\2x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}6x=-7\\3x=2\\2x=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-7}{6}\\x=\dfrac{2}{3}\\x=\dfrac{9}{2}\end{matrix}\right.\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{\dfrac{-7}{6};\dfrac{2}{3};\dfrac{9}{2}\right\}\)

\(7)\left(3,4x-6,8\right)\left(15x+2,5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3,4x-6,8=0\\15x+2,5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3,4x=6,8\\15x=-2,5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{6}\end{matrix}\right.\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{2;\dfrac{1}{6}\right\}\)

\(8)\left(x-1\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(\text{nhận}\right)\\x^2=-1\left(\text{loại}\right)\end{matrix}\right.\)

\(\text{Vậy phương trình trên có tập nghiệm là }S=\left\{1\right\}\)