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Câu hỏi: Biết rằng phương trình ${{\log }_{3}}\left( {{3}^{x+1}}-1 \right)=2\text{x}+{{\log }_{\dfrac{1}{3}}}2$ có hai nghiệm ${{x}_{1}}$ và ${{x}_{2}}$. Hãy tính tổng $S={{27}^{{{x}_{1}}}}+{{27}^{{{x}_{2}}}}$.
A. $S=180$
B. $S=45$
C. $S=9$
D. $S=252$
A. $S=180$
B. $S=45$
C. $S=9$
D. $S=252$
Điều kiện: ${{3}^{x+1}}-1>0\Leftrightarrow x>-1$.
Phương trình tương đương với: ${{\log }_{3}}\left( {{3}^{x+1}}-1 \right)=2\text{x}-{{\log }_{3}}2\Leftrightarrow {{\log }_{3}}\left( {{3}^{x+1}}-1 \right)+{{\log }_{3}}2=2\text{x}$
$\Leftrightarrow {{\log }_{3}}\left[ \left( {{3}^{x+1}}-1 \right).2 \right]=2\text{x}\Leftrightarrow \left( {{3}^{x+1}}-1 \right).2={{3}^{2\text{x}}}\Leftrightarrow {{6.3}^{x}}-2={{3}^{2\text{x}}}$
$\Leftrightarrow {{3}^{2\text{x}}}-{{6.3}^{x}}+2=0\xrightarrow{Vi-et}\left\{ \begin{aligned}
& {{3}^{{{x}_{1}}}}+{{3}^{{{x}_{2}}}}=6 \\
& {{3}^{{{x}_{1}}}}{{.3}^{{{x}_{2}}}}=2 \\
\end{aligned} \right.$
Ta có $S={{27}^{{{x}_{1}}}}+{{27}^{{{x}_{2}}}}={{\left( {{3}^{{{x}_{1}}}}+{{3}^{{{x}_{2}}}} \right)}^{3}}-{{3.3}^{{{x}_{1}}}}{{.3}^{{{x}_{2}}}}.\left( {{3}^{{{x}_{1}}}}+{{3}^{{{x}_{2}}}} \right)={{6}^{3}}-3.2.6=180$.
Phương trình tương đương với: ${{\log }_{3}}\left( {{3}^{x+1}}-1 \right)=2\text{x}-{{\log }_{3}}2\Leftrightarrow {{\log }_{3}}\left( {{3}^{x+1}}-1 \right)+{{\log }_{3}}2=2\text{x}$
$\Leftrightarrow {{\log }_{3}}\left[ \left( {{3}^{x+1}}-1 \right).2 \right]=2\text{x}\Leftrightarrow \left( {{3}^{x+1}}-1 \right).2={{3}^{2\text{x}}}\Leftrightarrow {{6.3}^{x}}-2={{3}^{2\text{x}}}$
$\Leftrightarrow {{3}^{2\text{x}}}-{{6.3}^{x}}+2=0\xrightarrow{Vi-et}\left\{ \begin{aligned}
& {{3}^{{{x}_{1}}}}+{{3}^{{{x}_{2}}}}=6 \\
& {{3}^{{{x}_{1}}}}{{.3}^{{{x}_{2}}}}=2 \\
\end{aligned} \right.$
Ta có $S={{27}^{{{x}_{1}}}}+{{27}^{{{x}_{2}}}}={{\left( {{3}^{{{x}_{1}}}}+{{3}^{{{x}_{2}}}} \right)}^{3}}-{{3.3}^{{{x}_{1}}}}{{.3}^{{{x}_{2}}}}.\left( {{3}^{{{x}_{1}}}}+{{3}^{{{x}_{2}}}} \right)={{6}^{3}}-3.2.6=180$.
Đáp án A.