\(\Leftrightarrow\left(5x-7\right)\left(5x+7-x-3\right)=0\)
\(\Leftrightarrow\left(5x-7\right)\left(4x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=-1\end{matrix}\right.\)
\(25x^2-49=\left(5x-7\right)\left(x+3\right)\)
\(\Rightarrow\left(5x\right)^2-7^2=\left(5x-7\right)\left(x+3\right)\)
\(\Rightarrow\left(5x-7\right)\left(5x+7\right)-\left(5x-7\right)\left(x+3\right)=0\)
\(\Rightarrow\left(5x-7\right)\left(5x+7-x-3\right)=0\)
\(\Rightarrow\left(5x-7\right)\left(4x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x-7=0\\4x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=7\\4x=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=-1\end{matrix}\right.\)
vậy \(x\in\left\{\dfrac{7}{5};-1\right\}\)
tích cho mk nhé
