Question

x²-4x+3 ≥0

Answer 1

`x^2-4x+3>=0`

`<=>(x-3)(x-1)>=0`

TH1: `{(x-3>=0),(x-1>=0):} <=> {(x>=3),(x>=1):} <=>x>=3`

TH2: `{(x-3<=0),(x-1<=0):}<=>{(x<=3),(x<=1):}<=>x<=1`

Vậy `x>=3 ; x<=1`.

Answer 2

Ta có: \(x^2-4x+3\ge0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1\le0\\x-3\ge0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le1\\x\ge3\end{matrix}\right.\)