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Câu hỏi: Với mọi số thực dương a và b thoả mãn ${{a}^{2}}+{{b}^{2}}=8ab,$ mệnh đề nào dưới đây đúng?
A. $\log \left( a+b \right)=\dfrac{1}{2}\left( \log a+\log b \right)$
B. $\log \left( a+b \right)=\dfrac{1}{2}\left( 1+\log a+\log b \right)$
C. $\log \left( a+b \right)=1+\log a+\log b$
D. $\log \left( a+b \right)=\dfrac{1}{2}+\log a+\log b$
A. $\log \left( a+b \right)=\dfrac{1}{2}\left( \log a+\log b \right)$
B. $\log \left( a+b \right)=\dfrac{1}{2}\left( 1+\log a+\log b \right)$
C. $\log \left( a+b \right)=1+\log a+\log b$
D. $\log \left( a+b \right)=\dfrac{1}{2}+\log a+\log b$
Ta có ${{a}^{2}}+{{b}^{2}}=8ab\Leftrightarrow {{a}^{2}}+2ab+{{b}^{2}}=10ab\Leftrightarrow {{(a+b)}^{2}}=10ab$
$\Rightarrow \log {{(a+b)}^{2}}=\log \left( 10ab \right)$
$\Rightarrow 2\log \left( a+b \right)=\left( 1+\log a+\log b \right)$
$\Rightarrow \log \left( a+b \right)=\dfrac{1}{2}\left( 1+\log a+\log b \right)$
$\Rightarrow \log {{(a+b)}^{2}}=\log \left( 10ab \right)$
$\Rightarrow 2\log \left( a+b \right)=\left( 1+\log a+\log b \right)$
$\Rightarrow \log \left( a+b \right)=\dfrac{1}{2}\left( 1+\log a+\log b \right)$
Đáp án B.