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Câu hỏi: Biết rằng phương trình ${{4}^{x}}-{{8.2}^{x}}+15=0$ có đúng hai nghiệm thực phân biệt ${{x}_{1}},{{x}_{2}}\left( {{x}_{1}}>{{x}_{2}} \right).$ Tính $S={{x}_{1}}+2{{x}_{2}}.$
A. $S={{\log }_{2}}15.$
B. $S={{\log }_{2}}45.$
C. $S={{\log }_{2}}75.$
D. $S={{\log }_{2}}135.$
A. $S={{\log }_{2}}15.$
B. $S={{\log }_{2}}45.$
C. $S={{\log }_{2}}75.$
D. $S={{\log }_{2}}135.$
${{4}^{x}}-{{8.2}^{x}}+15=0\Leftrightarrow {{\left( {{2}^{x}} \right)}^{2}}-{{8.2}^{x}}+15=0\Leftrightarrow \left[ \begin{aligned}
& {{2}^{x}}=3 \\
& {{2}^{x}}=5 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x={{\log }_{2}}3 \\
& x={{\log }_{2}}5 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& {{x}_{1}}={{\log }_{2}}5 \\
& {{x}_{2}}={{\log }_{2}}3 \\
\end{aligned} \right.$
$\Rightarrow S={{\log }_{2}}5+2{{\log }_{2}}3={{\log }_{2}}5+{{\log }_{2}}{{3}^{2}}={{\log }_{2}}\left( 5.9 \right)={{\log }_{2}}45$.
& {{2}^{x}}=3 \\
& {{2}^{x}}=5 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x={{\log }_{2}}3 \\
& x={{\log }_{2}}5 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& {{x}_{1}}={{\log }_{2}}5 \\
& {{x}_{2}}={{\log }_{2}}3 \\
\end{aligned} \right.$
$\Rightarrow S={{\log }_{2}}5+2{{\log }_{2}}3={{\log }_{2}}5+{{\log }_{2}}{{3}^{2}}={{\log }_{2}}\left( 5.9 \right)={{\log }_{2}}45$.
Đáp án B.