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Câu hỏi: Phương trình ${{\log }_{x}}2+{{\log }_{2}}x=\dfrac{5}{2}$ có hai nghiệm ${{x}_{1}},{{x}_{2}}\left( {{x}_{1}}<{{x}_{2}} \right)$. Khi đó tổng ${{x}^{2}}_{1}+{{x}_{2}}$ bằng
A. $\dfrac{9}{2}$.
B. $3$.
C. $6$.
D. $\dfrac{9}{4}$.
A. $\dfrac{9}{2}$.
B. $3$.
C. $6$.
D. $\dfrac{9}{4}$.
Điều kiện phương trình: $x>0,x\ne 1$.
${{\log }_{x}}2+{{\log }_{2}}x=\dfrac{5}{2}\Leftrightarrow \dfrac{1}{{{\log }_{2}}x}+{{\log }_{2}}x=\dfrac{5}{2}\Leftrightarrow {{\left( {{\log }_{2}}x \right)}^{2}}-\dfrac{5}{2}{{\log }_{2}}x+1=0\Leftrightarrow \left[ \begin{aligned}
& {{\log }_{2}}x=2 \\
& {{\log }_{2}}x=\dfrac{1}{2} \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x=4 \\
& x=\sqrt{2} \\
\end{aligned} \right. $Suy ra $ {{x}_{1}}=\sqrt{2},{{x}_{2}}=4$.
Suy ra ${{x}^{2}}_{1}+{{x}_{2}}=6$.
${{\log }_{x}}2+{{\log }_{2}}x=\dfrac{5}{2}\Leftrightarrow \dfrac{1}{{{\log }_{2}}x}+{{\log }_{2}}x=\dfrac{5}{2}\Leftrightarrow {{\left( {{\log }_{2}}x \right)}^{2}}-\dfrac{5}{2}{{\log }_{2}}x+1=0\Leftrightarrow \left[ \begin{aligned}
& {{\log }_{2}}x=2 \\
& {{\log }_{2}}x=\dfrac{1}{2} \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x=4 \\
& x=\sqrt{2} \\
\end{aligned} \right. $Suy ra $ {{x}_{1}}=\sqrt{2},{{x}_{2}}=4$.
Suy ra ${{x}^{2}}_{1}+{{x}_{2}}=6$.
Đáp án C.