Cho các số thực x, y thỏa mãn \({{2021}^{{{x}^{3}}+\frac{3}{2{{x}^{2}}}-\frac{3}{2}}}={{\log }_{\sqrt[2021]{2020}}}\left[ 2004-\left( y-11 \right)\sqrt{y+1} \right]\) với x > 0 và...

* Hướng dẫn giải

\({{2021}^{{{x}^{3}}+\frac{3}{2{{x}^{2}}}-\frac{3}{2}}}={{\log }_{\sqrt[2021]{2020}}}\left[ 2004-\left( y-11 \right)\sqrt{y+1} \right]\)

\(\Leftrightarrow {{2021}^{{{x}^{3}}+\frac{3}{2{{x}^{2}}}-\frac{3}{2}}}=2021{{\log }_{2020}}\left[ 2004-\left( y-11 \right)\sqrt{y+1} \right]\)

Ta có: \({{x}^{3}}+\frac{3}{2{{x}^{2}}}=\frac{{{x}^{3}}}{2}+\frac{{{x}^{3}}}{2}+\frac{1}{2{{x}^{2}}}+\frac{1}{2{{x}^{2}}}+\frac{1}{2{{x}^{2}}}\overset{cauchy}{\mathop{\ge }}\,\frac{5}{2},\forall x>0\Rightarrow VT\ge {{2021}^{\frac{5}{2}-\frac{3}{2}}}=2021\text{ }\left( 1 \right)\)

Ta có: \(2004-\left( y-11 \right)\sqrt{y+1}=2004-{{\left( \sqrt{y+1} \right)}^{3}}+12\sqrt{y+1}\)

Đặt \(t=\sqrt{y+1}\Rightarrow t\ge 0.\)

\(f\left( t \right)=2004-{{t}^{3}}+12t\)

\(\Rightarrow f'\left( t \right)=-3{{t}^{2}}+12\)

\(f'\left( t \right)=0\Leftrightarrow t=\pm 2.\)

Dựa vào BBT, ta có \(f\left( t \right)\le 2020,\) dấu “=” xảy ra \(\Leftrightarrow t=2.\)

\(\Rightarrow VP\le 2021.{{\log }_{2020}}2020=2021.1=2021\text{  }\left( 2 \right)\)

Từ \(\left( 1 \right)\) và \(\left( 2 \right)\Rightarrow \) Dấu “=” xảy ra đồng thời ở \(\left( 1 \right)\) và \(\left( 2 \right)\)

\(\Leftrightarrow \left\{ \begin{align} & \frac{{{x}^{3}}}{2}=\frac{1}{2{{x}^{2}}} \\ & \sqrt{y+1}=2 \\ \end{align} \right.\)

\(\Leftrightarrow \left\{ \begin{align} & x=1 \\ & y=3 \\ \end{align} \right.\Rightarrow P=11.\)