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Câu hỏi: Cho $2$ số thực $x$, $a$ với $x>a$ và $x>0$. Biết ${~\int\limits_{a}^{x}{\dfrac{f(t)}{{{t}^{2}}}\text{d}t}+6=2\sqrt{x}}$. Tìm $a$.
A. $29$.
B. $9$.
C. $19$.
D. $5$.
A. $29$.
B. $9$.
C. $19$.
D. $5$.
Ta có $\int\limits_{a}^{x}{\dfrac{f(t)}{{{t}^{2}}}\text{d}t}+6=2\sqrt{x}$ nên $\dfrac{f\left( x \right)}{{{x}^{2}}}=\dfrac{1}{\sqrt{x}}\Leftrightarrow f\left( x \right)=x\sqrt{x}$.
Do đó $~\int\limits_{a}^{x}{\dfrac{f(t)}{{{t}^{2}}}\text{d}t}=\int\limits_{a}^{x}{\dfrac{t\sqrt{t}}{{{t}^{2}}}\text{d}t}=\int\limits_{a}^{x}{{{t}^{-\dfrac{1}{2}}}\text{d}t}=\left. 2{{t}^{\dfrac{1}{2}}} \right|_{a}^{x}=2\sqrt{x}-2\sqrt{a}$.
Suy ra $~2\sqrt{x}-2\sqrt{a}+6=2\sqrt{x}\Leftrightarrow \sqrt{a}=3\Leftrightarrow a=9$.
Do đó $~\int\limits_{a}^{x}{\dfrac{f(t)}{{{t}^{2}}}\text{d}t}=\int\limits_{a}^{x}{\dfrac{t\sqrt{t}}{{{t}^{2}}}\text{d}t}=\int\limits_{a}^{x}{{{t}^{-\dfrac{1}{2}}}\text{d}t}=\left. 2{{t}^{\dfrac{1}{2}}} \right|_{a}^{x}=2\sqrt{x}-2\sqrt{a}$.
Suy ra $~2\sqrt{x}-2\sqrt{a}+6=2\sqrt{x}\Leftrightarrow \sqrt{a}=3\Leftrightarrow a=9$.
Đáp án B.