`x^2-4x+3>=0`
`<=>x^2-x-3x+3>=0`
`<=>x(x-1)-3(x-1)>=0`
`<=>(x-1)(x-3)>=0`
`<=>` \(\left[ \begin{array}{l}\begin{cases}x-1 \ge 0\\x-3 \ge 0\end{cases}\\\begin{cases}x-1 \le 0\\x-3 \le 0\end{cases}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}\begin{cases}x\ge 1\\x \ge 3\end{cases}\\\begin{cases}x \le 1\\x \le 3\end{cases}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x \ge 3\\x \le 1\end{array} \right.\)
`x^2-4x+3>=0`
`<=>(x^2-x)-(3x-3)>=0`
`<=>x(x-1)-3(x-1)>=0`
`<=>(x-1)(x-3)>=0`
TH1: \(\left\{{}\begin{matrix}x-1\ge0\\x-3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x\ge3\end{matrix}\right.\Leftrightarrow x\ge3\)
TH2: \(\left\{{}\begin{matrix}x-1\le0\\x-3\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x\le3\end{matrix}\right.\Leftrightarrow x\le1\)
Vậy `x<=1 ; 3 <= x` thỏa mãn.
\(x^2-4x+3\ge0\\ \Leftrightarrow\left(x-1\right)\left(x-3\right)\ge0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1\ge0\\x-3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1\le0\\x-3\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge1\\x\ge3\end{matrix}\right.\\\left\{{}\begin{matrix}x\le1\\x\le3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge3\\x\le1\end{matrix}\right.\)
Vậy \(S=\left\{x|x\ge3;x\le1\right\}\) là tập nghiệm của bất phương trình.
\(x^2-4x+3\ge0< =>\left(x-2\right)^2-1\ge0\)
\(< =>\left(x-2+1\right)\left(x-2-1\right)\ge0\)
\(< =>\left(x-1\right)\left(x-3\right)\ge0\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1\ge0\\x-3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1\le0\\x-3\le0\end{matrix}\right.\end{matrix}\right.< =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge1\\x\ge3\end{matrix}\right.\\\left\{{}\begin{matrix}x\le1\\x\le3\end{matrix}\right.\end{matrix}\right.\)\(< =>\left[{}\begin{matrix}x\ge3\\x\le1\end{matrix}\right.\)
vậy \(\left[{}\begin{matrix}x\ge3\\x\le1\end{matrix}\right.\) thì \(x^2-4x+3\ge0\)
