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Câu hỏi: Biết rằng đồ thị hàm số $y={{x}^{3}}-\left( 2a+1 \right){{x}^{2}}+\left( 2{{a}^{2}}+2a \right)x+b$ cắt trục hoành tại ba điểm phân biệt có hoành độ dương ${{x}_{1}},{{x}_{2}},{{x}_{3}}$. Tìm giá trị nhỏ nhất của biểu thức $P=x_{1}^{2}x_{2}^{3}x_{3}^{4}$.
A. $\min P=\dfrac{8\sqrt{3}}{729}$.
B. $\min P=\dfrac{64\sqrt{2}}{6561}$.
C. $\min P=\dfrac{32\sqrt{3}}{6561}$.
D. $\min P=\dfrac{2\sqrt{2}}{729}$.
A. $\min P=\dfrac{8\sqrt{3}}{729}$.
B. $\min P=\dfrac{64\sqrt{2}}{6561}$.
C. $\min P=\dfrac{32\sqrt{3}}{6561}$.
D. $\min P=\dfrac{2\sqrt{2}}{729}$.
Ta có $\left\{ \begin{aligned}
& {{x}_{1}}+{{x}_{2}}+{{x}_{3}}=2a+1 \\
& {{x}_{1}}{{x}_{2}}+{{x}_{2}}{{x}_{3}}+{{x}_{3}}{{x}_{1}}=2{{a}^{2}}+2a \\
\end{aligned} \right.\Rightarrow {{\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}} \right)}^{2}}-2\left( {{x}_{1}}{{x}_{2}}+{{x}_{2}}{{x}_{3}}+{{x}_{3}}{{x}_{1}} \right)=1$.
Do vậy: $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=1$. Xét các số thực dương $p,q,r$ sao cho đẳng thức xảy ra khi ${{x}_{1}}=p$, ${{x}_{2}}=q$, ${{x}_{3}}=r$.
Áp dụng AM – GM: $\dfrac{2{{x}_{1}}}{p}+\dfrac{3{{x}_{2}}}{q}+\dfrac{4{{x}_{3}}}{r}=\dfrac{{{x}_{1}}}{p}+\dfrac{{{x}_{1}}}{p}+\dfrac{{{x}_{2}}}{q}+\dfrac{{{x}_{2}}}{q}+\dfrac{{{x}_{2}}}{q}+\dfrac{{{x}_{3}}}{r}+\dfrac{{{x}_{3}}}{r}+\dfrac{{{x}_{3}}}{r}+\dfrac{{{x}_{3}}}{r}\ge 9\sqrt[9]{\dfrac{x_{1}^{2}x_{2}^{3}x_{3}^{4}}{{{p}^{2}}{{q}^{3}}{{r}^{4}}}}$.
Lại có: ${{\left( \dfrac{2{{x}_{1}}}{p}+\dfrac{3{{x}_{2}}}{q}+\dfrac{4{{x}_{3}}}{r} \right)}^{2}}\le \left( x_{1}^{2}+x_{2}^{2}+x_{3}^{2} \right)\left( \dfrac{4}{{{p}^{2}}}+\dfrac{9}{{{q}^{2}}}+\dfrac{16}{{{r}^{2}}} \right)=\dfrac{4}{{{p}^{2}}}+\dfrac{9}{{{q}^{2}}}+\dfrac{16}{{{r}^{2}}}$.
Khi đó ta có đẳng thức xảy ra khi: ${{x}_{1}}:{{x}_{2}}:{{x}_{3}}=\dfrac{2}{p}:\dfrac{3}{q}:\dfrac{4}{r}\Leftrightarrow \dfrac{p{{x}_{1}}}{2}=\dfrac{q{{x}_{2}}}{3}=\dfrac{r{{x}_{3}}}{4}\Leftrightarrow \dfrac{{{p}^{2}}}{2}=\dfrac{{{q}^{2}}}{3}=\dfrac{{{r}^{2}}}{4}$.
Mà ${{p}^{2}}+{{q}^{2}}+{{r}^{2}}=1$ nên $p=\dfrac{\sqrt{2}}{3};q=\dfrac{\sqrt{3}}{3};r=\dfrac{2}{3}$ do đó: $\dfrac{2{{x}_{1}}}{p}+\dfrac{3{{x}_{2}}}{q}+\dfrac{4{{x}_{3}}}{r}\le 9$ nên $\sqrt[9]{\dfrac{x_{1}^{2}x_{2}^{3}x_{3}^{4}}{{{p}^{2}}{{q}^{3}}{{r}^{4}}}}\le 1$.
Vậy: $x_{1}^{2}x_{2}^{3}x_{3}^{4}\le {{p}^{2}}{{q}^{3}}{{r}^{4}}=\dfrac{32\sqrt{3}}{6561}$ nên $\min P=\dfrac{32\sqrt{3}}{6561}$.
& {{x}_{1}}+{{x}_{2}}+{{x}_{3}}=2a+1 \\
& {{x}_{1}}{{x}_{2}}+{{x}_{2}}{{x}_{3}}+{{x}_{3}}{{x}_{1}}=2{{a}^{2}}+2a \\
\end{aligned} \right.\Rightarrow {{\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}} \right)}^{2}}-2\left( {{x}_{1}}{{x}_{2}}+{{x}_{2}}{{x}_{3}}+{{x}_{3}}{{x}_{1}} \right)=1$.
Do vậy: $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=1$. Xét các số thực dương $p,q,r$ sao cho đẳng thức xảy ra khi ${{x}_{1}}=p$, ${{x}_{2}}=q$, ${{x}_{3}}=r$.
Áp dụng AM – GM: $\dfrac{2{{x}_{1}}}{p}+\dfrac{3{{x}_{2}}}{q}+\dfrac{4{{x}_{3}}}{r}=\dfrac{{{x}_{1}}}{p}+\dfrac{{{x}_{1}}}{p}+\dfrac{{{x}_{2}}}{q}+\dfrac{{{x}_{2}}}{q}+\dfrac{{{x}_{2}}}{q}+\dfrac{{{x}_{3}}}{r}+\dfrac{{{x}_{3}}}{r}+\dfrac{{{x}_{3}}}{r}+\dfrac{{{x}_{3}}}{r}\ge 9\sqrt[9]{\dfrac{x_{1}^{2}x_{2}^{3}x_{3}^{4}}{{{p}^{2}}{{q}^{3}}{{r}^{4}}}}$.
Lại có: ${{\left( \dfrac{2{{x}_{1}}}{p}+\dfrac{3{{x}_{2}}}{q}+\dfrac{4{{x}_{3}}}{r} \right)}^{2}}\le \left( x_{1}^{2}+x_{2}^{2}+x_{3}^{2} \right)\left( \dfrac{4}{{{p}^{2}}}+\dfrac{9}{{{q}^{2}}}+\dfrac{16}{{{r}^{2}}} \right)=\dfrac{4}{{{p}^{2}}}+\dfrac{9}{{{q}^{2}}}+\dfrac{16}{{{r}^{2}}}$.
Khi đó ta có đẳng thức xảy ra khi: ${{x}_{1}}:{{x}_{2}}:{{x}_{3}}=\dfrac{2}{p}:\dfrac{3}{q}:\dfrac{4}{r}\Leftrightarrow \dfrac{p{{x}_{1}}}{2}=\dfrac{q{{x}_{2}}}{3}=\dfrac{r{{x}_{3}}}{4}\Leftrightarrow \dfrac{{{p}^{2}}}{2}=\dfrac{{{q}^{2}}}{3}=\dfrac{{{r}^{2}}}{4}$.
Mà ${{p}^{2}}+{{q}^{2}}+{{r}^{2}}=1$ nên $p=\dfrac{\sqrt{2}}{3};q=\dfrac{\sqrt{3}}{3};r=\dfrac{2}{3}$ do đó: $\dfrac{2{{x}_{1}}}{p}+\dfrac{3{{x}_{2}}}{q}+\dfrac{4{{x}_{3}}}{r}\le 9$ nên $\sqrt[9]{\dfrac{x_{1}^{2}x_{2}^{3}x_{3}^{4}}{{{p}^{2}}{{q}^{3}}{{r}^{4}}}}\le 1$.
Vậy: $x_{1}^{2}x_{2}^{3}x_{3}^{4}\le {{p}^{2}}{{q}^{3}}{{r}^{4}}=\dfrac{32\sqrt{3}}{6561}$ nên $\min P=\dfrac{32\sqrt{3}}{6561}$.
Đáp án C.