Đáp án:
\[m > \frac{3}{2}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
y < 0,\,\,\,\,\forall x \in \left( {1;2} \right)\\
\Leftrightarrow {x^2} - 2\left( {m + 1} \right)x + 4 < 0,\,\,\,\,\forall x \in \left( {1;2} \right)\\
\Leftrightarrow \left( {{x^2} - 2x + 4} \right) - 2mx < 0,\,\,\,\,\forall x \in \left( {1;2} \right)\\
\Leftrightarrow 2mx > {x^2} - 2x + 4,\,\,\,\,\forall x \in \left( {1;2} \right)\\
\Leftrightarrow m > \frac{{{x^2} - 2x + 4}}{{2x}},\,\,\,\,\forall x \in \left( {1;2} \right)\\
\Leftrightarrow m > \mathop {\max }\limits_{\left( {1;2} \right)} \frac{{{x^2} - 2x + 4}}{{2x}} = \frac{3}{2}\\
\Leftrightarrow m > \frac{3}{2}
\end{array}\)
