a) \(2x\left(3x-1\right)=3x-1\)
\(\Leftrightarrow2x\left(3x-1\right)-\left(3x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=0\\3x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x=1\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy x=\(\left\{\dfrac{1}{3};\dfrac{1}{2}\right\}\)
b) \(3\left(x-5\right)\left(x+2\right)=x^2-5x\)
\(\Leftrightarrow3\left(x^2-5x+2x-10\right)=x^2-5x\)
\(\Leftrightarrow3x^2-9x-30=x^2-5x\)
\(\Leftrightarrow3x^2-x^2-9x+5x-30=0\)
\(\Leftrightarrow2x^2-4x=30\)
\(\Leftrightarrow2x\left(x-4\right)=30\) (sau đó lập bảng và tìm giá trị của x)
c) \(\left(x-1\right)\left(2x+3\right)+2x=2\)
\(\Leftrightarrow2x^2+3x-2x-3+2x-2=0\)
\(\Leftrightarrow2x^2+3x-5=0\)
\(\Leftrightarrow x\left(2x+3\right)=5\)
Ta có bẳng sau
x...............-5...............-1...............1...............5
2x+3........-1...............-5...............5...............1
2x............-4...............-8...............2...............-2
x..............-2...............-4...............1...............-1
Vậy...