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Câu hỏi: Biết $\int_1^2\left(\sqrt[3]{x-\dfrac{1}{x^2}}+2 \sqrt[3]{\dfrac{1}{x^8}-\dfrac{1}{x^{11}}}\right) d x=\dfrac{a}{b} \sqrt[3]{c}$ với $a, b, c$ nguyên dương, $\dfrac{a}{b}$ tối giản và $c<a$. Tính $S=a+b-c$
A. 67 .
B. 75 .
C. $S=51$.
D. $S=39$.
A. 67 .
B. 75 .
C. $S=51$.
D. $S=39$.
$
\begin{aligned}
& \text { Xét } \left.I=\int_1^2\left(\sqrt[3]{x-\dfrac{1}{x^2}}+2 \sqrt[3]{\dfrac{1}{x^8}-\dfrac{1}{x^{11}}}\right) d x=\int_1^2\left(\sqrt[3]{x-\dfrac{1}{x^2}}+2^3 \sqrt[3]{\dfrac{1}{x^9}\left(x-\dfrac{1}{x^2}\right.}\right)\right) d x \\
& =\int_1^2\left(\sqrt[3]{x-\dfrac{1}{x^2}}+2 \cdot \dfrac{1}{x^3} \sqrt[3]{x-\dfrac{1}{x^2}}\right) d x=\int_1^2\left[\left(1+\dfrac{2}{x^3}\right)^{\sqrt[3]{x-\dfrac{1}{x^2}}}\right] d x
\end{aligned}
$
Đặt $t=\sqrt[3]{x-\dfrac{1}{x^2}} \Rightarrow t^3=x-\dfrac{1}{x^2} \Rightarrow 3 t^2 d t=\left(1+\dfrac{2}{x^3}\right) d x$.
Đổi cận $x=1 \Rightarrow t=0 ; x=2 \Rightarrow t=\sqrt[3]{\dfrac{7}{4}}$.
Vậy $I=\int_0^{\sqrt[3]{\dfrac{7}{4}}} t .3 t^2 d t=3 \int_0^{\sqrt[3]{\dfrac{7}{4}}} t^3 d t=\left.\dfrac{3}{4} t^4\right|_0 ^{\sqrt[3]{\dfrac{7}{4}}}=\dfrac{21}{16} \sqrt[3]{\dfrac{7}{4}}=\dfrac{21}{32} \sqrt[3]{14}$
Từ đó ta suy ra $a=21 ; b=32 ; c=14 \Rightarrow S=a+b-c=39$.
\begin{aligned}
& \text { Xét } \left.I=\int_1^2\left(\sqrt[3]{x-\dfrac{1}{x^2}}+2 \sqrt[3]{\dfrac{1}{x^8}-\dfrac{1}{x^{11}}}\right) d x=\int_1^2\left(\sqrt[3]{x-\dfrac{1}{x^2}}+2^3 \sqrt[3]{\dfrac{1}{x^9}\left(x-\dfrac{1}{x^2}\right.}\right)\right) d x \\
& =\int_1^2\left(\sqrt[3]{x-\dfrac{1}{x^2}}+2 \cdot \dfrac{1}{x^3} \sqrt[3]{x-\dfrac{1}{x^2}}\right) d x=\int_1^2\left[\left(1+\dfrac{2}{x^3}\right)^{\sqrt[3]{x-\dfrac{1}{x^2}}}\right] d x
\end{aligned}
$
Đặt $t=\sqrt[3]{x-\dfrac{1}{x^2}} \Rightarrow t^3=x-\dfrac{1}{x^2} \Rightarrow 3 t^2 d t=\left(1+\dfrac{2}{x^3}\right) d x$.
Đổi cận $x=1 \Rightarrow t=0 ; x=2 \Rightarrow t=\sqrt[3]{\dfrac{7}{4}}$.
Vậy $I=\int_0^{\sqrt[3]{\dfrac{7}{4}}} t .3 t^2 d t=3 \int_0^{\sqrt[3]{\dfrac{7}{4}}} t^3 d t=\left.\dfrac{3}{4} t^4\right|_0 ^{\sqrt[3]{\dfrac{7}{4}}}=\dfrac{21}{16} \sqrt[3]{\dfrac{7}{4}}=\dfrac{21}{32} \sqrt[3]{14}$
Từ đó ta suy ra $a=21 ; b=32 ; c=14 \Rightarrow S=a+b-c=39$.
Đáp án D.