\(d,\Rightarrow\left(x-1\right)\left(2x+x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{2}{3}\end{matrix}\right.\\ e,\Rightarrow\left(2x+5\right)\left(x-3-4x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=-1\end{matrix}\right.\\ f,\Rightarrow\left(5-x\right)\left(4x-2x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ g,\Rightarrow\left(x-3\right)\left(x-4-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=5\end{matrix}\right.\)
d: Ta có: \(2x\left(x-1\right)+\left(x-2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{2}{3}\end{matrix}\right.\)
e: Ta có: \(\left(x-3\right)\left(2x+5\right)=4x\left(2x+5\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(-3x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=-1\end{matrix}\right.\)
