Trên đoạn [0;2019] có bao nhiêu số nguyên m để phương trình 9^x -2(m+2).3^x

* Hướng dẫn giải

Chọn D

$$\begin{gathered} \mathrm{Đặt} 3^{\mathrm{x}}=\mathrm{t}. \mathrm{PT}(*)\hspace{0.17em}\mathrm{trở} \mathrm{thành} {\mathrm{t}}^2-2(\mathrm{m}+2)\mathrm{t}+3\mathrm{m}-2=0 \left(1\right) \\ \mathrm{PT} (*) \mathrm{có} 2 \mathrm{nghiệm} \mathrm{phân} \mathrm{biệt} \mathrm{khi} \mathrm{PT}\left(1\right) \mathrm{có} 2 \mathrm{nghiệm} \mathrm{dương} \mathrm{phân} \mathrm{biệt} \\ \iff \left\{\begin{array}{l}∆\text{'}>0\\ \mathrm{S}>0\\ \mathrm{P}>0\end{array}\iff \left\{\begin{array}{l}{(\mathrm{m}+2)}^2+2-3\mathrm{m}>0\\ 2(\mathrm{m}+2)>0\\ 3\mathrm{m}-2>0\end{array}\right.\right. \\ \iff \left\{\begin{array}{l}{\mathrm{m}}^2+\mathrm{m}+6>0 (\mathrm{luôn} \mathrm{đúng})\\ \mathrm{m}>-2\\ \mathrm{m}>\frac{2}{3}\end{array}\iff \mathrm{m}>\frac{2}{3}\right. \\ \mathrm{Khi} \mathrm{đó}\left\{\begin{array}{l}{3}^{{\mathrm{x}}_1}={\mathrm{t}}_1\\ 3^{{\mathrm{x}}_2}={\mathrm{t}}_2\end{array}\right.\iff \left\{\begin{array}{l}{\mathrm{x}}_1={\log }_3{\mathrm{t}}_1\\ {\mathrm{x}}_2={\log }_3{\mathrm{t}}_2\end{array}\right. \\ \mathrm{Từ} \mathrm{ĐB} \Rightarrow {\mathrm{x}}_1{\mathrm{x}}_2<0\Rightarrow 0<{\mathrm{t}}_1<1<{\mathrm{t}}_2\iff ({\mathrm{t}}_1-1)({\mathrm{t}}_2-1)<0 \\ \iff {\mathrm{t}}_1{\mathrm{t}}_2-({\mathrm{t}}_1+{\mathrm{t}}_2)+1<0. \mathrm{Theo} \mathrm{Vi}-\mathrm{ét} \left\{\begin{array}{l}{\mathrm{t}}_1+{\mathrm{t}}_2=2(\mathrm{m}+2)\\ {\mathrm{t}}_1{\mathrm{t}}_2=3\mathrm{m}-2\end{array}\right. \\ \Rightarrow 3\mathrm{m}-2-2(\mathrm{m}+2)+1<0\iff \mathrm{m}-5<0\iff \mathrm{m}<5 \\ \mathrm{mà} \mathrm{m}\in \left[0;2019\right]\Rightarrow 0\le \mathrm{m}<5 \\ \mathrm{Vậy} \mathrm{có} 5 \mathrm{giá} \mathrm{trị} \mathrm{nguyên} \mathrm{của} \mathrm{m} \mathrm{thỏa} \mathrm{mãn} \mathrm{bài} \mathrm{toán} \end{gathered}$$