Google
Câu hỏi: Tích các nghiệm của phương trình ${{\log }_{3}}\left( 3x \right).{{\log }_{3}}\left( 9x \right)=4$ là
A. $\dfrac{1}{3}.$
B. $\dfrac{1}{27}.$
C. $\dfrac{4}{3}.$
D. 1.
A. $\dfrac{1}{3}.$
B. $\dfrac{1}{27}.$
C. $\dfrac{4}{3}.$
D. 1.
Ta có ${{\log }_{3}}\left( 3x \right).{{\log }_{3}}\left( 9x \right)=4\Leftrightarrow \left( 1+{{\log }_{3}}x \right)\left( 2+{{\log }_{3}}x \right)=4$
${{\left( {{\log }_{3}}x \right)}^{2}}+3{{\log }_{3}}x+2=4\Leftrightarrow {{\left( {{\log }_{3}}x \right)}^{2}}+3{{\log }_{3}}x-2=0$
$\Leftrightarrow \left[ \begin{array}{*{35}{l}}
{{\log }_{3}}x=\dfrac{-3+\sqrt{17}}{2} \\
{{\log }_{3}}x=\dfrac{-3-\sqrt{17}}{2} \\
\end{array} \right.$
Do đó ${{\log }_{3}}{{x}_{1}}+{{\log }_{3}}{{x}_{2}}=-3\Leftrightarrow {{\log }_{3}}\left( {{x}_{1}}{{x}_{2}} \right)=-3\Leftrightarrow {{x}_{1}}{{x}_{2}}=\dfrac{1}{27}.$ Chọn B
${{\left( {{\log }_{3}}x \right)}^{2}}+3{{\log }_{3}}x+2=4\Leftrightarrow {{\left( {{\log }_{3}}x \right)}^{2}}+3{{\log }_{3}}x-2=0$
$\Leftrightarrow \left[ \begin{array}{*{35}{l}}
{{\log }_{3}}x=\dfrac{-3+\sqrt{17}}{2} \\
{{\log }_{3}}x=\dfrac{-3-\sqrt{17}}{2} \\
\end{array} \right.$
Do đó ${{\log }_{3}}{{x}_{1}}+{{\log }_{3}}{{x}_{2}}=-3\Leftrightarrow {{\log }_{3}}\left( {{x}_{1}}{{x}_{2}} \right)=-3\Leftrightarrow {{x}_{1}}{{x}_{2}}=\dfrac{1}{27}.$ Chọn B
Đáp án B.