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Câu hỏi: Cho hai số thực dương $x,y$ thỏa mãn ${{\log }_{\sqrt{2}}}2021=1+{{\log }_{\sqrt{2}}}\left( \sqrt{1+{{x}^{2}}}+x \right)-{{\log }_{2}}\left( {{y}^{2}}-y\sqrt{{{y}^{2}}+2}+1 \right).$ Giá trị nhỏ nhất của biểu thức $P=x+y$ thuộc khoảng nào dưới đây?
A. $\left( 40;41 \right).$
B. $\left( 42;43 \right).$
C. $\left( 44;45 \right).$
D. $\left( 46;47 \right).$
A. $\left( 40;41 \right).$
B. $\left( 42;43 \right).$
C. $\left( 44;45 \right).$
D. $\left( 46;47 \right).$
${{\log }_{\sqrt{2}}}2021=1+{{\log }_{\sqrt{2}}}\left( \sqrt{1+{{x}^{2}}}+x \right)-{{\log }_{2}}\left( {{y}^{2}}-y\sqrt{{{y}^{2}}+2}+1 \right)$
$\Leftrightarrow {{\log }_{2}}{{2021}^{2}}+{{\log }_{2}}\left( {{y}^{2}}-y\sqrt{{{y}^{2}}+2}+1 \right)={{\log }_{2}}2+{{\log }_{2}}{{\left( \sqrt{1+{{x}^{2}}}+x \right)}^{2}}$
$\Leftrightarrow {{2021}^{2}}\left( {{y}^{2}}-y\sqrt{{{y}^{2}}+2}+1 \right)=2{{\left( \sqrt{1+{{x}^{2}}}+x \right)}^{2}}$
$\Leftrightarrow {{2021}^{2}}\left( 2{{y}^{2}}-2y\sqrt{{{y}^{2}}+2}+2 \right)=4{{\left( \sqrt{1+{{x}^{2}}}+x \right)}^{2}}$
$\Leftrightarrow {{2021}^{2}}{{\left( \sqrt{{{y}^{2}}+2}-y \right)}^{2}}=4{{\left( \sqrt{1+{{x}^{2}}}+x \right)}^{2}}$
$\Leftrightarrow 2021\left( \sqrt{{{y}^{2}}+2}-y \right)=2\left( \sqrt{1+{{x}^{2}}}+x \right)\Rightarrow \left\{ \begin{aligned}
& \left( \sqrt{1+{{x}^{2}}}+x \right)\left( \sqrt{1+{{y}^{2}}}+y \right)=2021\left( 1 \right) \\
& \left( \sqrt{1+{{x}^{2}}}-x \right)\left( \sqrt{{{y}^{2}}+2}-y \right)=\dfrac{2}{2021}\left( 2 \right) \\
\end{aligned} \right.$
Công 2 vế (1) và (2) ta được:
$2021+\dfrac{2}{2021}=2\sqrt{\left( 1+{{x}^{2}} \right)\left( {{y}^{2}}+2 \right)}+2xy\le 1+{{x}^{2}}+{{y}^{2}}+2+2xy={{\left( x+y \right)}^{2}}+3$
$\Rightarrow x+y\ge \sqrt{2021+\dfrac{2}{2021}-3}\approx 44,922.$
$\Leftrightarrow {{\log }_{2}}{{2021}^{2}}+{{\log }_{2}}\left( {{y}^{2}}-y\sqrt{{{y}^{2}}+2}+1 \right)={{\log }_{2}}2+{{\log }_{2}}{{\left( \sqrt{1+{{x}^{2}}}+x \right)}^{2}}$
$\Leftrightarrow {{2021}^{2}}\left( {{y}^{2}}-y\sqrt{{{y}^{2}}+2}+1 \right)=2{{\left( \sqrt{1+{{x}^{2}}}+x \right)}^{2}}$
$\Leftrightarrow {{2021}^{2}}\left( 2{{y}^{2}}-2y\sqrt{{{y}^{2}}+2}+2 \right)=4{{\left( \sqrt{1+{{x}^{2}}}+x \right)}^{2}}$
$\Leftrightarrow {{2021}^{2}}{{\left( \sqrt{{{y}^{2}}+2}-y \right)}^{2}}=4{{\left( \sqrt{1+{{x}^{2}}}+x \right)}^{2}}$
$\Leftrightarrow 2021\left( \sqrt{{{y}^{2}}+2}-y \right)=2\left( \sqrt{1+{{x}^{2}}}+x \right)\Rightarrow \left\{ \begin{aligned}
& \left( \sqrt{1+{{x}^{2}}}+x \right)\left( \sqrt{1+{{y}^{2}}}+y \right)=2021\left( 1 \right) \\
& \left( \sqrt{1+{{x}^{2}}}-x \right)\left( \sqrt{{{y}^{2}}+2}-y \right)=\dfrac{2}{2021}\left( 2 \right) \\
\end{aligned} \right.$
Công 2 vế (1) và (2) ta được:
$2021+\dfrac{2}{2021}=2\sqrt{\left( 1+{{x}^{2}} \right)\left( {{y}^{2}}+2 \right)}+2xy\le 1+{{x}^{2}}+{{y}^{2}}+2+2xy={{\left( x+y \right)}^{2}}+3$
$\Rightarrow x+y\ge \sqrt{2021+\dfrac{2}{2021}-3}\approx 44,922.$
Đáp án C.