Câu hỏi của Hanz Zan

1: \(\left(x+1\right)^2\cdot\left(x+2\right)=0\)=>\(\left[{}\begin{matrix}\left(x+1\right)^2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)4: \(\left(3x-1\right)\left(3-x\right)^2=0\)=>\(\left[{}\begin{matrix}3x-1=0\\\left(3-x\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\3-x=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\end{matrix}\right.\)Câu trả lời: 1: \(\left(x+1\right)^2\cdot\left(x+2\right)=0\)=>\(\left[{}\begin{matrix}\left(x+1\right)^2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)4: \(\left(3x-1\right)\left(3-x\right)^2=0\)=>\(\left[{}\begin{matrix}3x-1=0\\\left(3-x\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\3-x=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\end{matrix}\right.\)