a) Ta có: \(x\left(x-2\right)-x^2+3x=4\)
\(\Leftrightarrow x^2-2x-x^2+3x=4\)
\(\Leftrightarrow x=4\)
Vậy: x=4
b) Ta có: \(3x^2-3x=\left(x-1\right)^2\)
\(\Leftrightarrow3x\left(x-1\right)-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\frac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{1;-\frac{1}{2}\right\}\)
c) Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-2\right)^2=-12\)
\(\Leftrightarrow x^3+8-x\left(x^2-4x+4\right)+12=0\)
\(\Leftrightarrow x^3+8-x^3+4x^2-4x+12=0\)
\(\Leftrightarrow4x^2-4x+20=0\)
\(\Leftrightarrow4x^2-2\cdot2x\cdot1+1+19=0\)
\(\Leftrightarrow\left(2x-1\right)^2+19=0\)(Vô lý)
Vậy: \(x\in\varnothing\)
