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Câu hỏi: Cho hàm số $f\left( x \right)={{e}^{\sqrt{1+\dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{\left( x+1 \right)}^{2}}}}}}$. Biết $f\left( 1 \right).f\left( 2 \right).f\left( 3 \right)...f\left( 2020 \right)={{e}^{\dfrac{m}{n}}}$ với $\dfrac{m}{n}$ là phân số tối giản. Tính $m-{{n}^{2}}$.
A. $-2020.$
B. 2020.
C. 1.
D. $-1.$
A. $-2020.$
B. 2020.
C. 1.
D. $-1.$
Ta có $1+\dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{\left( x+1 \right)}^{2}}}={{\left( \dfrac{1}{x}-\dfrac{1}{x+1} \right)}^{2}}+\dfrac{2}{x\left( x+1 \right)}+1={{\left( \dfrac{1}{x}-\dfrac{1}{x+1} \right)}^{2}}+2\left( \dfrac{1}{x}-\dfrac{1}{x+1} \right)+1={{\left( \dfrac{1}{x}-\dfrac{1}{x+1}+1 \right)}^{2}}$
Với $x>0\Rightarrow \dfrac{1}{x}>\dfrac{1}{x+1}\Rightarrow \sqrt{1+\dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{\left( x+1 \right)}^{2}}}}=\dfrac{1}{x}-\dfrac{1}{x+1}+1\Rightarrow f\left( x \right)=e.{{e}^{\dfrac{1}{x}-\dfrac{1}{x+1}}}$
Khi đó $f\left( 1 \right).f\left( 2 \right).f\left( 3 \right)...f\left( 2020 \right)={{e}^{2020}}.{{e}^{1-\dfrac{1}{2}}}.{{e}^{\dfrac{1}{2}-\dfrac{1}{3}}}.{{e}^{\dfrac{1}{3}-\dfrac{1}{4}}}...{{e}^{\dfrac{1}{2020}-\dfrac{1}{2021}}}={{e}^{2020}}.{{e}^{1-\dfrac{1}{2021}}}={{e}^{2021-\dfrac{1}{2021}}}$.
$\Rightarrow m={{2021}^{2}}-1;n=2021\Rightarrow m-{{n}^{2}}=-1$.
Với $x>0\Rightarrow \dfrac{1}{x}>\dfrac{1}{x+1}\Rightarrow \sqrt{1+\dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{\left( x+1 \right)}^{2}}}}=\dfrac{1}{x}-\dfrac{1}{x+1}+1\Rightarrow f\left( x \right)=e.{{e}^{\dfrac{1}{x}-\dfrac{1}{x+1}}}$
Khi đó $f\left( 1 \right).f\left( 2 \right).f\left( 3 \right)...f\left( 2020 \right)={{e}^{2020}}.{{e}^{1-\dfrac{1}{2}}}.{{e}^{\dfrac{1}{2}-\dfrac{1}{3}}}.{{e}^{\dfrac{1}{3}-\dfrac{1}{4}}}...{{e}^{\dfrac{1}{2020}-\dfrac{1}{2021}}}={{e}^{2020}}.{{e}^{1-\dfrac{1}{2021}}}={{e}^{2021-\dfrac{1}{2021}}}$.
$\Rightarrow m={{2021}^{2}}-1;n=2021\Rightarrow m-{{n}^{2}}=-1$.
Đáp án D.