\(x\left(3x-4\right)-6x+8=0\Leftrightarrow x\left(3x-4\right)-2\left(3x-4\right)\Leftrightarrow\left(3x-4\right)\left(x-2\right)\Leftrightarrow\left\{{}\begin{matrix}3x-4=0\\x-2=0\end{matrix}\right.\left\{{}\begin{matrix}x=\frac{4}{3}\\x=2\end{matrix}\right.\)Vậy: S = \(\left\{\frac{4}{3};2\right\}\)
x ( 3x - 4 ) - 6x + 8= 0
\(\Leftrightarrow\)x (3x - 4 ) - 2 (3x - 4 ) = 0
\(\Leftrightarrow\) ( 3x - 4 ).( x - 2 ) = 0
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3x-4=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=\frac{4}{3}\\x=2\end{matrix}\right.\)
Vậy S=\(\left\{\frac{4}{3};2\right\}\)
a, x(3x-4)-6x+8=0
\(\Leftrightarrow x\left(3x-4\right)-\left(6x-8\right)\)\(=0\)
\(\Leftrightarrow x\left(3x-4\right)-2\left(3x-4\right)\)\(=0\)
\(\Leftrightarrow\left(3x-4\right)\left(x-2\right)=0\)
\(\Leftrightarrow\) \(3x-4=0\) hoặc \(x-2=0\)
\(\Leftrightarrow\)\(3x=4\) hoặc \(x-2=0\)
\(\Leftrightarrow\) \(x=\frac{4}{3}\) hoặc \(x=2\)
Vậy phương trình có 2 nghiệm là \(x=\frac{4}{3}\) ; \(x=2\)
