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Câu hỏi: Cho hai số thực $x, y$ thỏa mãn : $9{{x}^{3}}+\left( 2-y\sqrt{3xy-5} \right)x+\sqrt{3xy-5}=0$. Tìm giá trị nhỏ nhất của $P={{x}^{3}}+{{y}^{3}}+6xy+3\left( 3{{x}^{2}}+1 \right)\left( x+y-2 \right)$ ?
A. $\dfrac{296\sqrt{15}-18}{9}$.
B. $\dfrac{36+296\sqrt{15}}{9}$.
C. $\dfrac{-4\sqrt{6}+18}{9}$.
D. $\dfrac{36-4\sqrt{6}}{9}$.
A. $\dfrac{296\sqrt{15}-18}{9}$.
B. $\dfrac{36+296\sqrt{15}}{9}$.
C. $\dfrac{-4\sqrt{6}+18}{9}$.
D. $\dfrac{36-4\sqrt{6}}{9}$.
Ta có: $9{{x}^{3}}+\left( 2-y\sqrt{3xy-5} \right)x+\sqrt{3xy-5}=0$ $\Leftrightarrow 9{{x}^{3}}+2x-xy\sqrt{3xy-5}+\sqrt{3xy-5}=0$
$\Leftrightarrow 27{{x}^{3}}+6x-3xy\sqrt{3xy-5}+3\sqrt{3xy-5}=0$ $\Leftrightarrow 27{{x}^{3}}+6x-\left( 3xy-5+5 \right)\sqrt{3xy-5}+3\sqrt{3xy-5}=0$
$\Leftrightarrow {{\left( 3x \right)}^{3}}+2.3x={{\left( \sqrt{3xy-5} \right)}^{3}}+2.\sqrt{3xy-5}$ $\left( 1 \right)$
Xét hàm số $f\left( t \right)={{t}^{3}}+2t$, có $f'\left( t \right)=3{{t}^{2}}+2>0,\forall t.$ Hay hàm $f\left( t \right)$ đồng biến trên $\mathbb{R}$.
Từ $\left( 1 \right)$ suy ra $f\left( 3x \right)=f\left( \sqrt{3xy-5} \right)\Leftrightarrow 3x=\sqrt{3xy-5}$
$\Leftrightarrow \left\{ \begin{aligned}
& x\ge 0 \\
& 9{{x}^{2}}=3xy-5 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x>0; y>0 \\
& 3xy=9{{x}^{2}}+5 \\
\end{aligned} \right.$$\left( 2 \right)$.
Xét $P={{x}^{3}}+{{y}^{3}}+6xy+3\left( 3{{x}^{2}}+1 \right)\left( x+y-2 \right)$
$\begin{aligned}
& P={{\left( x+y \right)}^{3}}-3xy\left( x+y \right)+6xy+\left( 9{{x}^{2}}+3 \right)\left( x+y-2 \right) \\
& P={{\left( x+y \right)}^{3}}-3xy\left( x+y \right)+6xy+\left( 3xy-2 \right)\left[ \left( x+y \right)-2 \right] \\
\end{aligned}$
$P={{\left( x+y \right)}^{3}}-2\left( x+y \right)+4$
Đặt $t=x+y\overset{do \left( 2 \right)}{\mathop{=}} x+\dfrac{9{{x}^{2}}+5}{3x}=4x+\dfrac{5}{3x}\ge 2\sqrt{4x.\dfrac{5}{3x}}\ge \dfrac{4\sqrt{15}}{3}.$
Xét hàm số $g\left( t \right)={{t}^{3}}-2t+4, t\ge \dfrac{4\sqrt{15}}{3}.$.Có $g'\left( t \right)=3{{t}^{2}}-2>0,\forall t\ge \dfrac{4\sqrt{15}}{3}$.
Do đó ${{P}_{\min }}=g\left( \dfrac{4\sqrt{15}}{3} \right)=\dfrac{36+296\sqrt{15}}{9}$.
$\Leftrightarrow 27{{x}^{3}}+6x-3xy\sqrt{3xy-5}+3\sqrt{3xy-5}=0$ $\Leftrightarrow 27{{x}^{3}}+6x-\left( 3xy-5+5 \right)\sqrt{3xy-5}+3\sqrt{3xy-5}=0$
$\Leftrightarrow {{\left( 3x \right)}^{3}}+2.3x={{\left( \sqrt{3xy-5} \right)}^{3}}+2.\sqrt{3xy-5}$ $\left( 1 \right)$
Xét hàm số $f\left( t \right)={{t}^{3}}+2t$, có $f'\left( t \right)=3{{t}^{2}}+2>0,\forall t.$ Hay hàm $f\left( t \right)$ đồng biến trên $\mathbb{R}$.
Từ $\left( 1 \right)$ suy ra $f\left( 3x \right)=f\left( \sqrt{3xy-5} \right)\Leftrightarrow 3x=\sqrt{3xy-5}$
$\Leftrightarrow \left\{ \begin{aligned}
& x\ge 0 \\
& 9{{x}^{2}}=3xy-5 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x>0; y>0 \\
& 3xy=9{{x}^{2}}+5 \\
\end{aligned} \right.$$\left( 2 \right)$.
Xét $P={{x}^{3}}+{{y}^{3}}+6xy+3\left( 3{{x}^{2}}+1 \right)\left( x+y-2 \right)$
$\begin{aligned}
& P={{\left( x+y \right)}^{3}}-3xy\left( x+y \right)+6xy+\left( 9{{x}^{2}}+3 \right)\left( x+y-2 \right) \\
& P={{\left( x+y \right)}^{3}}-3xy\left( x+y \right)+6xy+\left( 3xy-2 \right)\left[ \left( x+y \right)-2 \right] \\
\end{aligned}$
$P={{\left( x+y \right)}^{3}}-2\left( x+y \right)+4$
Đặt $t=x+y\overset{do \left( 2 \right)}{\mathop{=}} x+\dfrac{9{{x}^{2}}+5}{3x}=4x+\dfrac{5}{3x}\ge 2\sqrt{4x.\dfrac{5}{3x}}\ge \dfrac{4\sqrt{15}}{3}.$
Xét hàm số $g\left( t \right)={{t}^{3}}-2t+4, t\ge \dfrac{4\sqrt{15}}{3}.$.Có $g'\left( t \right)=3{{t}^{2}}-2>0,\forall t\ge \dfrac{4\sqrt{15}}{3}$.
Do đó ${{P}_{\min }}=g\left( \dfrac{4\sqrt{15}}{3} \right)=\dfrac{36+296\sqrt{15}}{9}$.
Đáp án B.