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Câu hỏi: Cho số thực $a$ dương tuỳ ý. Đặt ${{a}^{\dfrac{5}{4}}}\sqrt{a\sqrt[3]{a}}={{a}^{p}}$. Khẳng định đúng là:
A. $p=\dfrac{19}{12}.$
B. $p=\dfrac{23}{12}.$
C. $p=\dfrac{13}{12}.$
D. $p=\dfrac{23}{24}.$
A. $p=\dfrac{19}{12}.$
B. $p=\dfrac{23}{12}.$
C. $p=\dfrac{13}{12}.$
D. $p=\dfrac{23}{24}.$
Ta có: ${{a}^{p}}={{a}^{\dfrac{5}{4}}}\sqrt{a\sqrt[3]{a}}={{a}^{\dfrac{5}{4}}}\sqrt{a.{{a}^{\dfrac{1}{3}}}}={{a}^{\dfrac{5}{4}}}\sqrt{{{a}^{\dfrac{4}{3}}}}={{a}^{\dfrac{5}{4}}}.{{a}^{\dfrac{2}{3}}}={{a}^{\dfrac{23}{12}}}$.
Đáp án B.