Cho các số thực a, b, x, y thỏa mãn a > 1, b > 1 và a^2x = b^2y = căn bậc hai của ab

* Hướng dẫn giải

Theo bài ra ta có:

$a^{2x}=b^{2y}=\sqrt{ab}$

$\iff \left\{\begin{array}{l}2x={\log }_a\sqrt{ab}=\frac{1}{2}+\frac{1}{2}{\log }_{a}b\\ 2y={\log }_b\sqrt{ab}=\frac{1}{2}+\frac{1}{2}{\log }_{b}a\end{array}\right.$

$\iff \left\{\begin{array}{l}x=\frac{1}{4}+\frac{1}{4}.\frac{1}{{\log }_{b}a}\\ y=x=\frac{1}{4}+\frac{1}{4}.{\log }_{b}a\end{array}\right.$

Đặt $t={\log }_{b}a,$ vì $a>1,b>1\Rightarrow t={\log }_{b}a>{\log }_{b}1=0$ ta có: $\left\{\begin{array}{l}x=\frac{1}{4}+\frac{1}{4}.\frac{1}{t}\\ y=\frac{1}{4}+\frac{1}{4}.t\end{array}\right.\left(t>0\right)$

Khi đó ta có:

$P=6x+y^2=6\left(\frac{1}{4}+\frac{1}{4}.\frac{1}{t}\right)+{\left(\frac{1}{4}+\frac{1}{4}t\right)}^2$

$P=\frac{3}{2}+\frac{3}{2}.\frac{1}{t}+\frac{1}{16}+\frac{1}{8}t+\frac{1}{16}{t}^2$

$P=\frac{1}{16}{t}^2+\frac{1}{8}t+\frac{3}{2t}+\frac{25}{16}\left(t>0\right)$

Ta có

$P\text{'}=\frac{1}{8}t+\frac{1}{8}-\frac{3}{2t^2}=\frac{t^3+t^2-12}{8t^2}$

$P\text{'}=0\iff t^3+t^2-12=0\iff t=2\left(tm\right)$

BBT:

Vậy $P_{\min }=P\left(2\right)=\frac{45}{16}.$

Chọn D.