Đáp án:
`2000` giá trị nguyên $m$
Giải thích các bước giải:
\(\begin{array}{l}
\quad (10x)^{\displaystyle{m + \dfrac{\log x}{10}}} \geqslant 10^{\displaystyle{\dfrac{11}{10}\log x}}\qquad (x>0)\\
\Leftrightarrow \left(m + \dfrac{\log x}{10}\right)\log(10x) \geqslant \dfrac{11}{10}\log x\\
\Leftrightarrow \left(m + \dfrac{\log x}{10}\right)\left(1 + \log x\right) \geqslant \dfrac{11}{10}\log x\\
\Leftrightarrow \left(10m + \log x\right)\left(1 + \log x\right) \geqslant 11\log x\\
\Leftrightarrow 10m\left(1 + \log x\right) + \log^2x + \log x\geqslant 11\log x\\
\Leftrightarrow m \geqslant \dfrac{10\log x - \log^2x}{10 + 10\log x}\qquad (*)\\
\text{Đặt}\ t = \log x\\
x\in (1;100) \Rightarrow t\in (0;2)\\
(*)\Leftrightarrow m\geqslant \dfrac{10t - t^2}{10 + 10t}\\
\text{Xét}\ f(t) = \dfrac{10t - t^2}{10 + 10t}\\
\Rightarrow f'(t) = - \dfrac{t^2 +2t - 10}{(10 + 10t)^2}\\
f'(t) = 0 \Leftrightarrow \left[\begin{array}{l}t = -1 + \sqrt{11}\\t = -1 - \sqrt{11}\end{array}\right.\\
\text{Bảng xét dấu:}\\
\begin{array}{c|ccc}t&-\infty&&-1 -\sqrt{11}&&-1&&0&&2&&-1 + \sqrt{11}&&+\infty\\\hline
f'(t)&&-&0&+&\ \ \Vert&+&\vert&+&\vert&+&0&-&
\end{array}\\
\text{Dựa vào bảng xét dấu ta được:}\\
\text{$f(t)$ đồng biến trên $(0;2)$}\\
\Rightarrow \mathop{\max}\limits_{(0;2)}f(t) = f(2) = \dfrac{8}{15}\\
\text{Bất phương trình nghiệm đúng $\forall x\in (1;100)$}\\
\Leftrightarrow m \geqslant \mathop{\max}\limits_{(0;2)}\dfrac{10t - t^2}{10 + 10t}\\
\Leftrightarrow m \geqslant \dfrac{8}{15}\\
\text{mà}\ m\in [-2000;2000];\ m\in \Bbb Z\\
\text{nên}\ m \in \underbrace{\{1;2;3;\dots;1999;2000\}}_{\text{2000 giá trị m}}
\end{array}\)
