a. \(x^2-4x+3>0\)
\(\Leftrightarrow x^2-x-3x+3>0\)
\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)>0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)>0\)
T/h1: \(\left\{{}\begin{matrix}x-1>0\\x-3>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>1\\x>3\end{matrix}\right.\)\(\Leftrightarrow x>3\)
T/h2: \(\left\{{}\begin{matrix}x-1< 0\\x-3< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 1\\x< 3\end{matrix}\right.\)\(\Leftrightarrow x< 1\)
Vậy S={x/x>3 hoặc x<1}
b. \(x^2-4x< 0\)
\(\Leftrightarrow x\left(x-4\right)< 0\)
T/h1: \(\left\{{}\begin{matrix}x>0\\x-4< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\x< 4\end{matrix}\right.\Leftrightarrow0< x< 4\)
T/h2: \(\left\{{}\begin{matrix}x< 0\\x-4>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 0\\x>4\end{matrix}\right.\) (loại)
Vậy S={x/0<x<4}
Giải:
a) \(x^2-4x+3>0\)
\(\Leftrightarrow x^2-4x+4-1>0\)
\(\Leftrightarrow\left(x-4\right)^2-1>0\)
\(\Leftrightarrow\left(x-3\right)\left(x-5\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3>0\\x-5>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3< 0\\x-5< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>3\\x>5\end{matrix}\right.\\\left\{{}\begin{matrix}x< 3\\x< 5\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>5\\x< 3\end{matrix}\right.\)
Vậy ...
b) \(x^2-4x< 0\)
\(\Leftrightarrow x\left(x-4\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x-4>0\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x-4< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x>4\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x< 4\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\0< x< 4\end{matrix}\right.\)
Vậy ...
