\(\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x+7=0\\x-5=0\\5x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{7}{2}\\x=5\\x=-\frac{1}{5}\end{matrix}\right.\)
Vậy...
Ta có : \(\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)
=> \(\left[{}\begin{matrix}2x+7=0\\x-5=0\\5x+1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}2x=-7\\x=5\\5x=-1\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-\frac{7}{2}\\x=5\\x=-\frac{1}{5}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{-\frac{7}{2},5,-\frac{1}{5}\right\}.\)
