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Câu hỏi: Với các số $a,\ b>0$ thỏa mãn ${{a}^{2}}+{{b}^{2}}=7ab$, biểu thức ${{\log }_{3}}\left( a+b \right)$ bằng
A. $\dfrac{1}{2}\left( 1+{{\log }_{3}}a+{{\log }_{3}}b \right)$.
B. $1+\dfrac{1}{2}\left( {{\log }_{3}}a+{{\log }_{3}}b \right)$.
C. $\dfrac{1}{2}\left( 3+{{\log }_{3}}a+{{\log }_{3}}b \right)$
D. $2+\dfrac{1}{2}\left( {{\log }_{3}}a+{{\log }_{3}}b \right)$.
A. $\dfrac{1}{2}\left( 1+{{\log }_{3}}a+{{\log }_{3}}b \right)$.
B. $1+\dfrac{1}{2}\left( {{\log }_{3}}a+{{\log }_{3}}b \right)$.
C. $\dfrac{1}{2}\left( 3+{{\log }_{3}}a+{{\log }_{3}}b \right)$
D. $2+\dfrac{1}{2}\left( {{\log }_{3}}a+{{\log }_{3}}b \right)$.
Ta có: ${{a}^{2}}+{{b}^{2}}=7ab\Leftrightarrow {{a}^{2}}+{{b}^{2}}+2ab=9ab\Leftrightarrow {{\left( a+b \right)}^{2}}=9ab$
Vì $a,b>0,$ ta có logarit 2 vế: $\log {{\left( a+b \right)}^{2}}={{\log }_{3}}\left( 9ab \right)$
$\Leftrightarrow 2{{\log }_{3}}\left( a+b \right)={{\log }_{3}}9+{{\log }_{3}}a+{{\log }_{3}}b$
$\Leftrightarrow {{\log }_{3}}\left( a+b \right)=\dfrac{1}{2}{{\log }_{3}}9+\dfrac{1}{2}\left( {{\log }_{3}}a+{{\log }_{3}}b \right)$
$\Leftrightarrow {{\log }_{3}}\left( a+b \right)=1+\dfrac{1}{2}\left( {{\log }_{3}}a+{{\log }_{3}}b \right)$
Vì $a,b>0,$ ta có logarit 2 vế: $\log {{\left( a+b \right)}^{2}}={{\log }_{3}}\left( 9ab \right)$
$\Leftrightarrow 2{{\log }_{3}}\left( a+b \right)={{\log }_{3}}9+{{\log }_{3}}a+{{\log }_{3}}b$
$\Leftrightarrow {{\log }_{3}}\left( a+b \right)=\dfrac{1}{2}{{\log }_{3}}9+\dfrac{1}{2}\left( {{\log }_{3}}a+{{\log }_{3}}b \right)$
$\Leftrightarrow {{\log }_{3}}\left( a+b \right)=1+\dfrac{1}{2}\left( {{\log }_{3}}a+{{\log }_{3}}b \right)$
Đáp án B.