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Câu hỏi: Cho số thực $x$ thỏa mãn ${{2}^{{{x}^{2}}}}{{.3}^{x+1}}=1.$ Mệnh đề nào dưới đây đúng?
A. ${{x}^{2}}+\left( x+1 \right){{\log }_{2}}3=0.$
B. ${{x}^{2}}+\left( x+1 \right){{\log }_{2}}3=1.$
C. $\left( x+1 \right)+{{x}^{2}}{{\log }_{3}}2=1.$
D. $\left( x+1 \right)+x{{\log }_{3}}2=0.$
A. ${{x}^{2}}+\left( x+1 \right){{\log }_{2}}3=0.$
B. ${{x}^{2}}+\left( x+1 \right){{\log }_{2}}3=1.$
C. $\left( x+1 \right)+{{x}^{2}}{{\log }_{3}}2=1.$
D. $\left( x+1 \right)+x{{\log }_{3}}2=0.$
${{2}^{{{x}^{2}}}}{{.3}^{x+1}}=1\Rightarrow {{\log }_{2}}\left( {{2}^{{{x}^{2}}}}{{.3}^{x+1}} \right)={{\log }_{2}}1\Leftrightarrow {{x}^{2}}+\left( x+1 \right){{\log }_{2}}3=0.$ Suy ra A đúng, B sai.
${{2}^{{{x}^{2}}}}{{.3}^{x+1}}=1\Rightarrow {{\log }_{3}}\left( {{2}^{{{x}^{2}}}}{{.3}^{x+1}} \right)={{\log }_{3}}1\Leftrightarrow {{x}^{2}}{{\log }_{3}}2+x+1=0.$ Suy ra C, D sai.
${{2}^{{{x}^{2}}}}{{.3}^{x+1}}=1\Rightarrow {{\log }_{3}}\left( {{2}^{{{x}^{2}}}}{{.3}^{x+1}} \right)={{\log }_{3}}1\Leftrightarrow {{x}^{2}}{{\log }_{3}}2+x+1=0.$ Suy ra C, D sai.
Đáp án A.