Đáp án: $B$
Giải thích các bước giải:
ĐKXĐ $x>2 ,y>3$
Ta có:
$\log_5(x-2)+\log_5(y-3)=1$
$\to \log_5(x-2)(y-3)=1$
$\to (x-2)(y-3)=5$
Ta có:
$P=3x+5y$
$\to P=3(x-2)+5(y-3)+21$
$\to P\ge 2\sqrt{3(x-2)\cdot 5(y-3)}+21$
$\to P\ge 2\sqrt{15(x-2)(y-3)}+21$
$\to P\ge 2\sqrt{15\cdot 5}+21$
$\to P\ge 10\sqrt{3}+21$
Dấu = xảy ra khi
$\begin{cases}3(x-2)=5(y-3)\\ (x-2)(y-3)=5\end{cases}$
$\to\begin{cases}(x-2)=\dfrac53(y-3)\\ \dfrac53(y-3)\cdot (y-3)=5\end{cases}$
$\to\begin{cases}(x-2)=\dfrac53(y-3)\\ (y-3)^2=3\end{cases}$
$\to\begin{cases}(x-2)=\dfrac53\cdot \sqrt{3}\\ y-3=\sqrt{3}, y-3>0\end{cases}$
$\to\begin{cases}x=2+\dfrac5{\sqrt{3}}\\ y=3+\sqrt{3}\end{cases}$
$\to 5x-3y=5(2+\dfrac5{\sqrt{3}})-3(3+\sqrt{3})=\dfrac{16\sqrt{3}}{3}+1$
$\to a=16, b=3\to a+2b=22\to T=22$
