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Câu hỏi: Cho hàm số $y=f\left( x \right)={{e}^{\dfrac{1}{x\left( x+1 \right)}}}$. Tính giá trị biểu thức $T=f\left( 1 \right).f\left( 2 \right)...f\left( 2017 \right).\sqrt[2018]{e}$
A. $T=1$
B. $T=e$
C. $T=\dfrac{1}{e}$
D. $T={{e}^{\dfrac{1}{2018}}}$
A. $T=1$
B. $T=e$
C. $T=\dfrac{1}{e}$
D. $T={{e}^{\dfrac{1}{2018}}}$
Ta có $f\left( x \right)={{e}^{\dfrac{1}{x\left( x+1 \right)}}}\Leftrightarrow f\left( x \right)=\dfrac{{{e}^{\dfrac{1}{x}}}}{{{e}^{\dfrac{1}{x+1}}}}$
$\Rightarrow T=f\left( 1 \right).f\left( 2 \right)...f\left( 2017 \right).\sqrt[2018]{e}=\dfrac{e}{{{e}^{\dfrac{1}{2}}}}.\dfrac{{{e}^{\dfrac{1}{2}}}}{{{e}^{\dfrac{1}{3}}}}...\dfrac{{{e}^{\dfrac{1}{2017}}}}{{{e}^{\dfrac{1}{2018}}}}.{{e}^{\dfrac{1}{2018}}}\Leftrightarrow T=e$
$\Rightarrow T=f\left( 1 \right).f\left( 2 \right)...f\left( 2017 \right).\sqrt[2018]{e}=\dfrac{e}{{{e}^{\dfrac{1}{2}}}}.\dfrac{{{e}^{\dfrac{1}{2}}}}{{{e}^{\dfrac{1}{3}}}}...\dfrac{{{e}^{\dfrac{1}{2017}}}}{{{e}^{\dfrac{1}{2018}}}}.{{e}^{\dfrac{1}{2018}}}\Leftrightarrow T=e$
Đáp án B.