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Câu hỏi: Cho biết phương trình ${{\log }_{3}}({{3}^{x+1}}-1)=2\text{x}+{{\log }_{\dfrac{1}{3}}}2$ có hai nghiệm ${{x}_{1}},{{x}_{2}}$. Hãy tính tổng $S={{27}^{{{x}_{1}}}}+{{27}^{{{x}_{2}}}}$.
A. $S=252$
B. $S=45$
C. $S=9$
D. $S=180$
A. $S=252$
B. $S=45$
C. $S=9$
D. $S=180$
${{3}^{x+1}}-1={{3}^{2\text{x}+{{\log }_{\dfrac{1}{3}}}2}}={{3}^{2x}}.\dfrac{1}{2}\Rightarrow {{3}^{x}}=t;\dfrac{1}{2}{{t}^{2}}-3t+1=0\Rightarrow {{3}^{x}}\in \left\{ 3-\sqrt{7};3+\sqrt{7} \right\}$
$\Rightarrow {{27}^{{{x}_{1}}}}+{{27}^{{{x}_{2}}}}={{\left( 3-\sqrt{7} \right)}^{{{x}_{1}}}}+{{\left( 3-\sqrt{7} \right)}^{{{x}_{2}}}}=180$.
$\Rightarrow {{27}^{{{x}_{1}}}}+{{27}^{{{x}_{2}}}}={{\left( 3-\sqrt{7} \right)}^{{{x}_{1}}}}+{{\left( 3-\sqrt{7} \right)}^{{{x}_{2}}}}=180$.
Đáp án D.