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Câu hỏi: Cho $x>0, y>1$ thỏa mãn $\dfrac{1}{2}{{y}^{2}}.{{\log }_{2}}\left( \dfrac{xy-x}{2y} \right)=-2{{\left( y-1 \right)}^{2}}+\dfrac{8{{y}^{2}}}{{{x}^{2}}}$. Giá trị nhỏ nhất của $P=\sqrt[4]{{{e}^{\dfrac{{{x}^{2}}}{1+2y}}}}. {{e}^{\dfrac{{{y}^{2}}}{x+1}}}$ có dạng ${{e}^{\dfrac{m}{n}}}$ (trong đó $m, n$ là các số nguyên dương, $\dfrac{m}{n}$ là phân số tối giản). Giá trị $m+n$ bằng
A. $12$.
B. $21$.
C. $22$.
D. $13$.
A. $12$.
B. $21$.
C. $22$.
D. $13$.
Với $x>0, y>1$, ta có:
$\dfrac{1}{2}{{y}^{2}}.{{\log }_{2}}\left( \dfrac{xy-x}{2y} \right)=-2{{\left( y-1 \right)}^{2}}+\dfrac{8{{y}^{2}}}{{{x}^{2}}}\Leftrightarrow {{\log }_{2}}\left( \dfrac{xy-x}{2y} \right)=-4{{\left( \dfrac{y-1}{y} \right)}^{2}}+\dfrac{16}{{{x}^{2}}}$
$\Leftrightarrow {{\log }_{2}}\dfrac{x}{2}-4{{\left( \dfrac{2}{x} \right)}^{2}}={{\log }_{2}}\left( \dfrac{y}{y-1} \right)-4{{\left( \dfrac{y-1}{y} \right)}^{2}} \left( * \right)$.
Xét hàm số $f\left( t \right)={{\log }_{2}}t-\dfrac{4}{{{t}^{2}}} \left( t>0 \right) \Rightarrow {f}'\left( t \right)=\dfrac{1}{t\ln 2}+\dfrac{8}{{{t}^{3}}}>0,\forall t>0\Rightarrow f\left( t \right)$ luôn đồng biến trên $\left( 0; +\infty \right)$. Khi đó $\left( * \right)$ có nghiệm $\dfrac{x}{2}=\dfrac{y}{y-1}\Leftrightarrow x=\dfrac{2y}{y-1}$.
Khi đó $P=\sqrt[4]{{{e}^{\dfrac{{{x}^{2}}}{1+2y}}}}. {{e}^{\dfrac{{{y}^{2}}}{x+1}}}={{e}^{\dfrac{{{\left( \dfrac{x}{2} \right)}^{2}}}{1+2y}+\dfrac{{{y}^{2}}}{x+1}}}$. Đặt $\dfrac{x}{2}=a>0 ; y=b>1$.
Từ $x=\dfrac{2y}{y-1}\Rightarrow \dfrac{x}{2}\left( y-1 \right)=y\Rightarrow a+b=ab$. Mặt khác, ta có:
${{\left( \dfrac{a+b}{2} \right)}^{2}}\ge ab\Rightarrow {{\left( \dfrac{a+b}{2} \right)}^{2}}\ge a+b\Rightarrow {{\left( a+b \right)}^{2}}-4\left( a+b \right)\ge 0\Rightarrow a+b\ge 4 \left( do a+b>0 \right).$
Ta có: $P={{e}^{\dfrac{{{a}^{2}}}{1+2b}+\dfrac{{{b}^{2}}}{1+2a}}}$. Theo bất đẳng thức BCS ta có: $\dfrac{{{a}^{2}}}{1+2b}+\dfrac{{{b}^{2}}}{1+2a}\ge \dfrac{{{\left( a+b \right)}^{2}}}{2+2\left( a+b \right)}$.
Xét hàm số $f\left( t \right)=\dfrac{{{t}^{2}}}{2+2t} , \left( t\ge 4 \right)\Rightarrow {f}'\left( t \right)=\dfrac{2{{t}^{2}}+4t}{{{\left( 2+2t \right)}^{2}}} >0,\forall t\ge 4\Rightarrow f\left( t \right)$ luôn đồng biến trên $\left[ 4; +\infty \right)$.
Suy ra $\underset{\left[ 4; +\infty \right)}{\mathop{\min }} f\left( t \right)=f\left( 4 \right)=\dfrac{8}{5}.$ Khi đó: ${{P}_{\min }}={{e}^{\dfrac{8}{5}}}={{e}^{\dfrac{m}{n}}}\Rightarrow m=8, n=5 \Rightarrow m+n=13.$.
$\dfrac{1}{2}{{y}^{2}}.{{\log }_{2}}\left( \dfrac{xy-x}{2y} \right)=-2{{\left( y-1 \right)}^{2}}+\dfrac{8{{y}^{2}}}{{{x}^{2}}}\Leftrightarrow {{\log }_{2}}\left( \dfrac{xy-x}{2y} \right)=-4{{\left( \dfrac{y-1}{y} \right)}^{2}}+\dfrac{16}{{{x}^{2}}}$
$\Leftrightarrow {{\log }_{2}}\dfrac{x}{2}-4{{\left( \dfrac{2}{x} \right)}^{2}}={{\log }_{2}}\left( \dfrac{y}{y-1} \right)-4{{\left( \dfrac{y-1}{y} \right)}^{2}} \left( * \right)$.
Xét hàm số $f\left( t \right)={{\log }_{2}}t-\dfrac{4}{{{t}^{2}}} \left( t>0 \right) \Rightarrow {f}'\left( t \right)=\dfrac{1}{t\ln 2}+\dfrac{8}{{{t}^{3}}}>0,\forall t>0\Rightarrow f\left( t \right)$ luôn đồng biến trên $\left( 0; +\infty \right)$. Khi đó $\left( * \right)$ có nghiệm $\dfrac{x}{2}=\dfrac{y}{y-1}\Leftrightarrow x=\dfrac{2y}{y-1}$.
Khi đó $P=\sqrt[4]{{{e}^{\dfrac{{{x}^{2}}}{1+2y}}}}. {{e}^{\dfrac{{{y}^{2}}}{x+1}}}={{e}^{\dfrac{{{\left( \dfrac{x}{2} \right)}^{2}}}{1+2y}+\dfrac{{{y}^{2}}}{x+1}}}$. Đặt $\dfrac{x}{2}=a>0 ; y=b>1$.
Từ $x=\dfrac{2y}{y-1}\Rightarrow \dfrac{x}{2}\left( y-1 \right)=y\Rightarrow a+b=ab$. Mặt khác, ta có:
${{\left( \dfrac{a+b}{2} \right)}^{2}}\ge ab\Rightarrow {{\left( \dfrac{a+b}{2} \right)}^{2}}\ge a+b\Rightarrow {{\left( a+b \right)}^{2}}-4\left( a+b \right)\ge 0\Rightarrow a+b\ge 4 \left( do a+b>0 \right).$
Ta có: $P={{e}^{\dfrac{{{a}^{2}}}{1+2b}+\dfrac{{{b}^{2}}}{1+2a}}}$. Theo bất đẳng thức BCS ta có: $\dfrac{{{a}^{2}}}{1+2b}+\dfrac{{{b}^{2}}}{1+2a}\ge \dfrac{{{\left( a+b \right)}^{2}}}{2+2\left( a+b \right)}$.
Xét hàm số $f\left( t \right)=\dfrac{{{t}^{2}}}{2+2t} , \left( t\ge 4 \right)\Rightarrow {f}'\left( t \right)=\dfrac{2{{t}^{2}}+4t}{{{\left( 2+2t \right)}^{2}}} >0,\forall t\ge 4\Rightarrow f\left( t \right)$ luôn đồng biến trên $\left[ 4; +\infty \right)$.
Suy ra $\underset{\left[ 4; +\infty \right)}{\mathop{\min }} f\left( t \right)=f\left( 4 \right)=\dfrac{8}{5}.$ Khi đó: ${{P}_{\min }}={{e}^{\dfrac{8}{5}}}={{e}^{\dfrac{m}{n}}}\Rightarrow m=8, n=5 \Rightarrow m+n=13.$.
Đáp án D.