\(\left(x-1\right)\left(2x+3\right)-x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+3-x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)
Vậy...
(x-1)(2x+3)-x(x-1)=0
2x2 +3x-2x-3-x2 +x=0
x2 +2x-3=0
x2 +2x=3
x(x+2)=3
Suy ra x+2=3
x=1
Vậy x=1
\(\left(x-1\right)\left(2x+3\right)-x\left(x-1\right)=0\)
\(\left(x-1\right)\left(2x+3-x\right)=0\)
\(\left(x-1\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{1\text{ ; }-3\right\}\)
