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Câu hỏi: Cho 2 số thực x, y thỏa mãn ${{\log }_{2}}\dfrac{{{x}^{2}}+{{y}^{2}}}{3xy+{{x}^{2}}}+{{x}^{2}}+2{{y}^{2}}+1\le 3xy$. Tìm giá trị nhỏ nhất của biểu thức $P=\dfrac{2{{x}^{2}}-xy+2{{y}^{2}}}{2xy-{{y}^{2}}}$.
A. $\dfrac{3}{2}$.
B. $\dfrac{5}{2}$.
C. $\dfrac{1}{2}$.
D. $\dfrac{7}{2}$.
A. $\dfrac{3}{2}$.
B. $\dfrac{5}{2}$.
C. $\dfrac{1}{2}$.
D. $\dfrac{7}{2}$.
Biến đổi giả thiết ta có:
${{\log }_{2}}\dfrac{{{x}^{2}}+{{y}^{2}}}{3xy+{{x}^{2}}}+1+2{{x}^{2}}+2{{y}^{2}}\le 3xy+{{x}^{2}}\Leftrightarrow {{\log }_{2}}\dfrac{2{{x}^{2}}+2{{y}^{2}}}{3xy+{{x}^{2}}}+2{{x}^{2}}+2{{y}^{2}}\le 3xy+{{x}^{2}}$
$\Leftrightarrow {{\log }_{2}}\left( 2{{x}^{2}}+2{{y}^{2}} \right)+2{{x}^{2}}+2{{y}^{2}}\le {{\log }_{2}}\left( 3xy+{{x}^{2}} \right)+3xy+{{x}^{2}}$
$\Leftrightarrow 2{{x}^{2}}+2{{y}^{2}}\le 3xy+{{x}^{2}}\Leftrightarrow {{x}^{2}}-3xy+2{{y}^{2}}\le 0\Leftrightarrow 1\le \dfrac{x}{y}\le 2$
Khi đó $P=\dfrac{2{{\left( \dfrac{x}{y} \right)}^{2}}-\dfrac{x}{y}+2}{\dfrac{2x}{y}-1}=f\left( \dfrac{x}{y} \right)\ge f\left( \dfrac{3}{2} \right)=\dfrac{5}{2}$.
${{\log }_{2}}\dfrac{{{x}^{2}}+{{y}^{2}}}{3xy+{{x}^{2}}}+1+2{{x}^{2}}+2{{y}^{2}}\le 3xy+{{x}^{2}}\Leftrightarrow {{\log }_{2}}\dfrac{2{{x}^{2}}+2{{y}^{2}}}{3xy+{{x}^{2}}}+2{{x}^{2}}+2{{y}^{2}}\le 3xy+{{x}^{2}}$
$\Leftrightarrow {{\log }_{2}}\left( 2{{x}^{2}}+2{{y}^{2}} \right)+2{{x}^{2}}+2{{y}^{2}}\le {{\log }_{2}}\left( 3xy+{{x}^{2}} \right)+3xy+{{x}^{2}}$
$\Leftrightarrow 2{{x}^{2}}+2{{y}^{2}}\le 3xy+{{x}^{2}}\Leftrightarrow {{x}^{2}}-3xy+2{{y}^{2}}\le 0\Leftrightarrow 1\le \dfrac{x}{y}\le 2$
Khi đó $P=\dfrac{2{{\left( \dfrac{x}{y} \right)}^{2}}-\dfrac{x}{y}+2}{\dfrac{2x}{y}-1}=f\left( \dfrac{x}{y} \right)\ge f\left( \dfrac{3}{2} \right)=\dfrac{5}{2}$.
Đáp án B.