Google
Câu hỏi: Cho biểu thức $P=\sqrt{x\sqrt[3]{{{x}^{2}}\sqrt[4]{{{x}^{3}}}}}$ với $x>0$. Mệnh đề nào dưới đây đúng?
A. $P={{x}^{\dfrac{13}{24}}}$.
B. $P={{x}^{\dfrac{23}{12}}}$.
C. $P={{x}^{\dfrac{12}{23}}}$
D. $P={{x}^{\dfrac{23}{24}}}$.
A. $P={{x}^{\dfrac{13}{24}}}$.
B. $P={{x}^{\dfrac{23}{12}}}$.
C. $P={{x}^{\dfrac{12}{23}}}$
D. $P={{x}^{\dfrac{23}{24}}}$.
Vì $x>0$ nên ta có:
$P=\sqrt{x\sqrt[3]{{{x}^{2}}\sqrt[4]{{{x}^{3}}}}}=\sqrt{x\sqrt[3]{{{x}^{2}}.{{x}^{\dfrac{3}{4}}}}}=\sqrt{x\sqrt[3]{{{x}^{^{\dfrac{11}{4}}}}}}=\sqrt{x.{{x}^{\dfrac{11}{12}}}}=\sqrt{{{x}^{\dfrac{23}{12}}}}={{x}^{\dfrac{23}{24}}}$.
Vậy $P={{x}^{\dfrac{23}{24}}}$.
$P=\sqrt{x\sqrt[3]{{{x}^{2}}\sqrt[4]{{{x}^{3}}}}}=\sqrt{x\sqrt[3]{{{x}^{2}}.{{x}^{\dfrac{3}{4}}}}}=\sqrt{x\sqrt[3]{{{x}^{^{\dfrac{11}{4}}}}}}=\sqrt{x.{{x}^{\dfrac{11}{12}}}}=\sqrt{{{x}^{\dfrac{23}{12}}}}={{x}^{\dfrac{23}{24}}}$.
Vậy $P={{x}^{\dfrac{23}{24}}}$.
Đáp án D.