a) Ta có: \(\left(2x-4\right)\left(3x+1\right)+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)\left[2\left(3x+1\right)+\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(6x+2+x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\cdot7x=0\)
Vì 7≠0
nên \(\left[{}\begin{matrix}x-2=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
Vậy: x∈{0;2}
b) Ta có: \(\left(2x+1\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(2x+1-x+1\right)\left(2x+1+x-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\cdot3x=0\)
Vì 3≠0
nên \(\left[{}\begin{matrix}x+2=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=0\end{matrix}\right.\)
Vậy: x∈{0;-2}
c) Ta có: \(2x^2-x=0\)
\(\Leftrightarrow x\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{1}{2}\right\}\)
d) Ta có: \(x^3-6x^2+9x=0\)
\(\Leftrightarrow x\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow x\left(x-3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy: x∈{0;3}
k) Ta có: \(x^3+3x^2+x+3=0\)
\(\Leftrightarrow x^2\left(x+3\right)+\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+1\right)=0\)(1)
Ta có: \(x^2+1\ge1>0\forall x\)(2)
Từ (1) và (2) suy ra x+3=0
hay x=-3
Vậy: x=-3