`(4x+1)(2x-5) =(2x-5)(2x+3)`
`<=> (4x+1)(2x-5)-(2x-5)(2x+3)=0`
`<=> (2x-5)[(4x+1)-(2x+3)]=0`
`<=> (2x-5)(4x+1-2x-3)=0`
`<=>(2x-5)(2x-2)=0`
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=1\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{5}{2};1\right\}\)
`(4x+1)(2x-5)=(2x-5)(2x+3)`
`<=>(4x+1)(2x-5)-(2x-5)(2x+3)=0`
`<=>(2x-5)[(4x+1-(2x+3)]=0`
`<=>(2x-5)(4x+1-2x-3)=0`
`<=>(2x-5)(2x-2)=0`
`<=>2x-5=0` hoặc `2x-2=0`
`<=>x=5/2` hoặc `x=1`
\(\left(4x+1\right)\left(2x-5\right)=\left(2x-5\right)\left(2x+3\right)\)
\(\Leftrightarrow8x^2-20x+2x-5=4x^2+6x-10x-15\)
\(\Leftrightarrow8x^2-4x^2-20x+2x-6x+10x=-15+5\)
\(\Leftrightarrow4x^2-14x=-10\)
\(\Leftrightarrow4x^2-14x+10=0\)
\(\left(2x-5\right)\left(2x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=1\end{matrix}\right.\)
