a: Ta có: \(2\left(x+1\right)=\left(5x-1\right)\left(x+1\right)\)
=>\(\left(5x-1\right)\left(x+1\right)-2\left(x+1\right)=0\)
=>(x+1)(5x-1-2)=0
=>(x+1)(5x-3)=0
=>\(\left[\begin{array}{l}x+1=0\\ 5x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-1\\ 5x=3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-1\\ x=\frac35\end{array}\right.\)
b: \(\left(-4x+3\right)\cdot x=\left(2x+5\right)\cdot x\)
=>\(x\left(-4x+3\right)-x\left(2x+5\right)=0\)
=>x(-4x+3-2x-5)=0
=>x(-6x-2)=0
=>-2x(3x+1)=0
=>x(3x+1)=0
=>\(\left[\begin{array}{l}x=0\\ 3x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ 3x=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-\frac13\end{array}\right.\)
