Xét các số thực x, y thỏa mãn log3(1 - y/x + 3xy) = 3xy + x + 3y - 4

Chọn B.

Điều kiện: $\frac{1-y}{x+3xy}>0.$ Vì x, y > 0 do đó $\frac{1-y}{x+3xy}>0\iff 1-y>0\iff 0<y<1$

Ta có:

${\log }_3\frac{1-y}{x+3xy}=3xy+x+3y-4\iff {\log }_3\left[3\left(1-y\right)\right]+3\left(1-y\right)={\log }_3\left(3xy+x\right)+\left(3xy+x\right)\left(1\right)$

Xét hàm số $f\left(t\right)={\log }_{3}t+t\left(t>0\right)$ ta có: $f\text{'}\left(t\right)=\frac{1}{t\ln 3}+1>0\left(\forall t>0\right).$

Suy ra hàm số f(t) đồng biến trên $\left(0;+\infty \right)$.

Suy ra:

$\left(1\right)\iff f\left(3\left(1-y\right)\right)=f\left(3xy+x\right)\iff 3\left(1-y\right)=3xy+x\iff x=\frac{3\left(1-y\right)}{3y+1}=\frac{4}{3y+1}-1\left(0<y<1\right)$

Suy ra $P=x+y=y+\frac{4}{3y+1}-1=\frac{1}{3}\left(3y+1\right)+\frac{4}{3y+1}-\frac{4}{3}\ge 2\sqrt{\frac{1}{3}\left(3y+1\right).\frac{4}{3y+1}}-\frac{4}{3}=\frac{4\sqrt{3}-4}{3}$

$\Rightarrow P_{\min }=\frac{4\sqrt{3}-4}{3}.$

Dấu “=” xảy ra $\iff \frac{1}{3}\left(3y+1\right)=\frac{4}{3y+1}\Rightarrow {\left(3y+1\right)}^2=12\iff \left[\begin{array}{l}y=\frac{2\sqrt{3}-1}{3}\left(TM\right)\Rightarrow x=\frac{2\sqrt{3}-3}{3}\\ y=\frac{-2\sqrt{3}-1}{3}\left(L\right)\end{array}\right.$.