| [tex]x^{2} - 4x + 3[/tex] | + |[tex]x^{2} -4x[/tex] | = 3
Ta có: VT $=|x^2-4x+3|+|x^2-4x|=|x^2-4x+3|+|4x-x^2|\ge |x^2-4x+3+4x-x^2|=3=$ VP.
Dấu '=' xảy ra $\Leftrightarrow \left\{\begin{matrix} x^2-4x+3\ge 0 \\ 4x-x^2\ge 0 \end{matrix} \right.\Leftrightarrow \left\{\begin{matrix} x\le 1 \ or \ x\ge 3 \\ 0\le x\le 4 \end{matrix} \right.\Leftrightarrow \left[\begin{matrix} 0\le x\le 1 \\ 3\le x\le 4 \end{matrix} \right.$
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