Cho x thỏa mãn phương trình

- Hướng dẫn giải

Đáp án là B.

Pt $\iff \left\{\begin{array}{l}{5.2}^x-8>0\\ \frac{{5.2}^x-8}{2^x+2}=2^{3-x}\end{array}\right.\iff \left\{\begin{array}{l}{2}^x>\frac{8}{5}\\ {5.2}^x-8=\frac{8}{2^x}\left(2^x+2\right)\end{array}\right.$

$\iff \left\{\begin{array}{l}x>{\log }_2\frac{8}{5}\\ {5.2}^{2x}-{16.2}^x-16=0\end{array}\right.$

$\iff \left\{\begin{array}{l}x>{\log }_2\frac{8}{5}\\ \left[\begin{array}{l}{2}^x=4\\ 2^x=-\frac{4}{5}\end{array}\right.\end{array}\right.\iff \left\{\begin{array}{l}x>{\log }_2\frac{8}{5}\\ x=2\end{array}\right.\Rightarrow x=2.$

$P=x^{{\log }_{2}4x}=2^{{\log }_{2}8}=8.$