Ta có: \(x^2-4x+3>0\)
<=> \(\left(x-2\right)^2-1>0\)
<=> \(\left(x-2-1\right)\left(x-2+1\right)>0\)
<=> \(\left(x-3\right)\left(x-1\right)>0\)
=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3>0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3< 0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>3\\x>1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 3\\x< 1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>3\\x< 1\end{matrix}\right.\)
Vậy x>3 hoặc x<1 là nghiệm của bất phương trình
\(x^2-4x+3>0\)
\(\Rightarrow\left(x-2\right)^2-1>0\)
\(\Rightarrow\left(x-2\right)^2>1\)
\(\Rightarrow\left[{}\begin{matrix}x-2>1\\x-2< -1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>3\\x< 1\end{matrix}\right.\)
Vậy x > 3 hoặc x < 1.
\(x^2-4x+3\ge0\)
\(\Leftrightarrow x^2-3x-x+3\ge0\Leftrightarrow x\left(x-3\right)-\left(x-3\right)\ge0\Leftrightarrow\left(x-1\right)\left(x-3\right)\ge0\Rightarrow\left[{}\begin{matrix}x-1\ge0\\x-3\ge0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x\ge1\\x\ge3\end{matrix}\right.\)
