Gọi S là tập hợp tất cả các giá trị nguyên của tham số m

* Hướng dẫn giải

Chọn B

$$\begin{gathered} {16}^x-m.4^{x-1}+5m^2-44=0 \iff {\left(4^x\right)}^2-\frac{m}{4}.4^x+5m^2-44=0 (*) \\ Đặt 4^x=t (t>0) PT(*)\hspace{0.17em}trở thành t^2-\frac{m}{4}t+5m^2-44=0 \\ \iff 4t^2-mt+4(5m^2-44)=0 \left(1\right) \\ P\mathrm{T}(*) \mathrm{có} 2 \mathrm{nghiệm} \mathrm{phân} \mathrm{biệt}\iff \mathrm{PT}\left(1\right) \mathrm{có} 2 \mathrm{nghiệm} \mathrm{dương} \mathrm{phân} \mathrm{biệt} \\ \iff \left\{\begin{array}{l}∆>0\\ \mathrm{S}>0\\ \mathrm{P}>0\end{array}\iff \left\{\begin{array}{l}{\mathrm{m}}^2-64(5{\mathrm{m}}^2-44)>0\\ \frac{\mathrm{m}}{4}>0\\ 5{\mathrm{m}}^2-44>0\end{array}\right.\right. \\ \iff \left\{\begin{array}{l}-319{\mathrm{m}}^2+2816>0\\ \mathrm{m}>0\\ {\mathrm{m}}^2>\frac{44}{5}\end{array}\iff \left\{\begin{array}{l}\frac{-16}{\sqrt{29}}<\mathrm{m}<\frac{16}{\sqrt{29}}\\ \mathrm{m}>0\\ \mathrm{m}\in \left(-\infty ;\frac{-2\sqrt{55}}{5}\right)\cup \left(\frac{2\sqrt{55}}{5};+\infty \right)\end{array}\right.\right. \\ \iff \frac{2\sqrt{55}}{5}<\mathrm{m}<\frac{16}{\sqrt{29}} \left(2,966<\mathrm{m}<2,971\right) \\ \mathrm{Khi} \mathrm{đó} \left\{\begin{array}{l}{4}^{{\mathrm{x}}_1}={\mathrm{t}}_1\\ 4^{{\mathrm{x}}_2}={\mathrm{t}}_2\end{array}\right.\iff \left\{\begin{array}{l}{\mathrm{x}}_1={\log }_4{\mathrm{t}}_1\\ {\mathrm{x}}_2={\log }_4{\mathrm{t}}_2\end{array}\right. \\ \mathrm{PT} \mathrm{có} 2 \mathrm{nghiệm} \mathrm{đối} \mathrm{nhau}\iff {\mathrm{x}}_1+{\mathrm{x}}_2=0 \\ \iff {\log }_4{\mathrm{t}}_1+ {\log }_4{\mathrm{t}}_2=0\iff {\log }_4\left({\mathrm{t}}_1{\mathrm{t}}_2\right)=0 \\ \iff {\log }_4(5{\mathrm{m}}^2-44)=0\iff 5{\mathrm{m}}^2-44=1 \\ \iff {\mathrm{m}}^2=9\iff \mathrm{m}=\pm 3 \left(\mathrm{L}\right) \\ \mathrm{KL}: \mathrm{Không} \mathrm{có} \mathrm{giá} \mathrm{trị} \mathrm{nào} \mathrm{của} \mathrm{m} \mathrm{TM} \mathrm{bài} \mathrm{toán} \end{gathered}$$