Cho 2 số thực dương x, y thỏa mãn

* Hướng dẫn giải

Với x, y > 0 ta có:

     ${\log }_5{\left[\left(x+2\right)\left(y+1\right)\right]}^{y+1}=125-\left(x-1\right)\left(y+1\right)$

$\iff \left(y+1\right)\left[{\log }_5\left(x+2\right)+{\log }_5\left(y+1\right)\right]=125-\left(x-1\right)\left(y+1\right)$

$\iff {\log }_5\left(x+2\right)+{\log }_5\left(y+1\right)=\frac{125}{y+1}-x+1$

$\iff {\log }_5\left(x+2\right)+{\log }_5\left(y+1\right)=\frac{125}{y+1}-\left(x+2\right)+3$

$\iff {\log }_5\left(x+2\right)+\left(x+2\right)={\log }_5\frac{125}{y+1}+\frac{125}{y+1}$

Xét hàm đặc trưng $f\left(t\right)={\log }_{5}t+t\left(t>0\right)$ ta có $f\text{'}\left(t\right)=\frac{1}{t\ln 5}+1>0\forall t>0$ suy ra hàm số đồng biến trên $\left(0;+\infty \right),$ do đó $f\left(x+2\right)=f\left(\frac{125}{y+1}\right)\iff x+2=\frac{125}{y+1}\iff x=\frac{125}{y+1}-2.$

Khi đó ta có $P=x+5y=\frac{125}{y+1}-2+5y.$

$\Rightarrow P=\frac{125}{y+1}+5\left(y+1\right)-7\ge 2\sqrt{\frac{125}{y+1}.5\left(y+1\right)}-7=43.$

Dấu “=” xảy ra khi $\frac{125}{y+1}=5\left(y+1\right)\iff {\left(y+1\right)}^2=25\iff \left[\begin{array}{l}y+1=5\\ y+1=-5\end{array}\right.\iff y=4$ (do y > 0).

Với $y=4\Rightarrow x=\frac{125}{5}-2=23.$

Vậy $P_{\min }=43\iff x=23,y=4.$

Chọn C.