Ta có :
\(7x^2-13x+6=0\)
=> \(7x^2-7x-13x+7x+6=0\)
=> \(7x\left(x-1\right)-6x+6=0\)
=>\(7x\left(x-1\right)-6\left(x-1\right)=0\)
=> \(\left(7x-6\right)\left(x-1\right)=0\)
=> \(\left[{}\begin{matrix}7x-6=0\\x-1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\dfrac{6}{7}\\x=1\end{matrix}\right.\)
Vậy x \(\in\)\(\left\{\dfrac{6}{7};1\right\}\)
\(7x^2-13x+6=0\Leftrightarrow7x^2-7x-6x+6=0\)
\(\Leftrightarrow7x\left(x-1\right)-6\left(x-1\right)=0\Leftrightarrow\left(7x-6\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\7x-6=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\7x=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=\dfrac{6}{7}\end{matrix}\right.\) vậy \(x=1;x=\dfrac{6}{7}\)
=> 7(x\(^2\)-\(\dfrac{13}{7}\)x + \(\dfrac{6}{7}\)) =0
=> x\(^2\)- 2. \(\dfrac{13}{14}\) + \(\dfrac{169}{196}\)- \(\dfrac{1}{196}\)=0
=>(x- \(\dfrac{13}{14}\))\(^2\) = \(\dfrac{1}{196}\)
=> \(\) x- \(\dfrac{13}{14}\)=\(\dfrac{1}{196}\) hoặc x- \(\dfrac{13}{14}\) = - \(\dfrac{1}{196}\)
=> x= \(\dfrac{1}{196}\)+\(\dfrac{13}{14}\) hoặc x = -\(\dfrac{1}{196}\)+\(\dfrac{13}{14}\)
