\(\left|x-2\right|+\left|3x^2-12\right|=0\)
Ta có:
\(\left\{{}\begin{matrix}\left|x-2\right|\ge0\\\left|3x^2-12\right|\ge0\end{matrix}\right.\forall x.\)
\(\Rightarrow\left|x-2\right|+\left|3x^2-12\right|\ge0\) \(\forall x.\)
\(\Rightarrow\left|x-2\right|+\left|3x^2-12\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-2\right|=0\\\left|3x^2-12\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-2=0\\3x^2-12=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\3x^2=12\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\x^2=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=2\\x=-2\left(loại\right)\end{matrix}\right.\Rightarrow x=2.\)
Vậy \(x=2.\)
Chúc bạn học tốt!
\(\left|x-2\right|+\left|3x^2-12\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-2\right|=0\\\left|3x^2-12\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-2=0\\3x^2-12=0\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=0+2\\3x^2=0+12\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=2\\3x^2=12\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=2\\x^2=12:3\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=2\\x^2=4\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=2\\\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x=2\)
