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Câu hỏi: Cho hàm số $y=\sqrt{x \sqrt{x \sqrt{x}}}$ xác định trên $(0 ;+\infty)$. Biết $y^{\prime}=m x^n$. Giá trị $m+n$ là
A. $\dfrac{9}{8}$.
B. $\dfrac{3}{4}$.
C. $\dfrac{9}{4}$.
D. $\dfrac{5}{8}$.
A. $\dfrac{9}{8}$.
B. $\dfrac{3}{4}$.
C. $\dfrac{9}{4}$.
D. $\dfrac{5}{8}$.
$
\begin{aligned}
& \text { Ta có } y=\sqrt{x \sqrt{x \sqrt{x}}}=\sqrt{x \sqrt{x \cdot x^{\dfrac{1}{2}}}}=\sqrt{x \sqrt{x^{\dfrac{3}{2}}}}=\sqrt{x \cdot x^{\dfrac{3}{4}}}=\sqrt{x^{\dfrac{7}{4}}}=x^{\dfrac{7}{8}} \\
& \Rightarrow m=\dfrac{7}{8}, n=-\dfrac{1}{8} \Rightarrow m+n=\dfrac{3}{4} \quad y^{\prime}=\dfrac{7}{8} x^{-\dfrac{1}{8}}
\end{aligned}
$
\begin{aligned}
& \text { Ta có } y=\sqrt{x \sqrt{x \sqrt{x}}}=\sqrt{x \sqrt{x \cdot x^{\dfrac{1}{2}}}}=\sqrt{x \sqrt{x^{\dfrac{3}{2}}}}=\sqrt{x \cdot x^{\dfrac{3}{4}}}=\sqrt{x^{\dfrac{7}{4}}}=x^{\dfrac{7}{8}} \\
& \Rightarrow m=\dfrac{7}{8}, n=-\dfrac{1}{8} \Rightarrow m+n=\dfrac{3}{4} \quad y^{\prime}=\dfrac{7}{8} x^{-\dfrac{1}{8}}
\end{aligned}
$
Đáp án B.