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Câu hỏi: Giải phương trình ${{\left( 27\sqrt{3} \right)}^{{{x}^{2}}-x+1}}={{9}^{x+1}}.$
A. $x=\dfrac{10\pm \sqrt{35}}{12}.$
B. $x=\dfrac{10\pm \sqrt{37}}{14}.$
C. $x=\dfrac{11\pm \sqrt{35}}{12}.$
D. $x=\dfrac{11\pm \sqrt{37}}{14}.$
A. $x=\dfrac{10\pm \sqrt{35}}{12}.$
B. $x=\dfrac{10\pm \sqrt{37}}{14}.$
C. $x=\dfrac{11\pm \sqrt{35}}{12}.$
D. $x=\dfrac{11\pm \sqrt{37}}{14}.$
Ta có ${{\left( 27\sqrt{3} \right)}^{{{x}^{2}}-x+1}}={{9}^{x+1}}\Leftrightarrow {{\left( {{3}^{3}}{{.3}^{\dfrac{1}{2}}} \right)}^{{{x}^{2}}-x+1}}={{\left( {{3}^{2}} \right)}^{x+1}}\Leftrightarrow {{\left( {{3}^{3+\dfrac{1}{2}}} \right)}^{{{x}^{2}}-x+1}}={{3}^{2\left( x+1 \right)}}$.
$\Leftrightarrow {{3}^{\dfrac{7}{2}\left( {{x}^{2}}-x+1 \right)}}={{3}^{2\left( x+1 \right)}}\Leftrightarrow \dfrac{7}{2}\left( {{x}^{2}}-x+1 \right)=2\left( x+1 \right)\Leftrightarrow x=\dfrac{11\pm \sqrt{37}}{14}.$
$\Leftrightarrow {{3}^{\dfrac{7}{2}\left( {{x}^{2}}-x+1 \right)}}={{3}^{2\left( x+1 \right)}}\Leftrightarrow \dfrac{7}{2}\left( {{x}^{2}}-x+1 \right)=2\left( x+1 \right)\Leftrightarrow x=\dfrac{11\pm \sqrt{37}}{14}.$
Đáp án D.