Yên Chi
[44616]
22/11/2018 12:47:39 PM
Phương trình logarit
Với $x\ne 1$ có \[{{\log }_{2}}\left( x-\sqrt{{{x}^{2}}-1} \right).{{\log }_{3}}\left( x+\sqrt{{{x}^{2}}-1} \right)={{\log }_{6}}\left( x-\sqrt{{{x}^{2}}-1} \right)\]
\[\Leftrightarrow {{\log }_{2}}\left( x-\sqrt{{{x}^{2}}-1} \right).{{\log }_{3}}\left( \dfrac{1}{x-\sqrt{{{x}^{2}}-1}} \right)={{\log }_{6}}\left( x-\sqrt{{{x}^{2}}-1} \right).\]
\[\Leftrightarrow -{{\log }_{2}}\left( x-\sqrt{{{x}^{2}}-1} \right).{{\log }_{3}}\left( x-\sqrt{{{x}^{2}}-1} \right)={{\log }_{6}}\left( x-\sqrt{{{x}^{2}}-1} \right).\]
\[\Leftrightarrow {{\log }_{3}}\left( x-\sqrt{{{x}^{2}}-1} \right)=-\frac{{{\log }_{6}}\left( x-\sqrt{{{x}^{2}}-1} \right)}{{{\log }_{2}}\left( x-\sqrt{{{x}^{2}}-1} \right)}=-{{\log }_{6}}2\Leftrightarrow x-\sqrt{{{x}^{2}}-1}={{3}^{-{{\log }_{6}}2}}.\]
\[\Leftrightarrow x-{{3}^{-{{\log }_{6}}2}}=\sqrt{{{x}^{2}}-1}\Leftrightarrow {{\left( x-{{3}^{-{{\log }_{6}}2}} \right)}^{2}}={{x}^{2}}-1.\]
\[\Leftrightarrow {{x}^{2}}-2x{{.3}^{-{{\log }_{6}}2}}+{{3}^{-2{{\log }_{6}}2}}={{x}^{2}}-1\Leftrightarrow 2x{{.3}^{-{{\log }_{6}}2}}={{3}^{-2{{\log }_{6}}2}}+1\Leftrightarrow x=\dfrac{1}{2}\left( {{3}^{-{{\log }_{6}}2}}+{{3}^{{{\log }_{6}}2}} \right).\]
Khi đó $a=3,b=6,c=2\Rightarrow {{a}^{2}}-2b+3c=9-12+6=3.$
Chọn đáp án B.