Cho các số thực dương x,y thỏa mãn (10/9)^(2x^2 - 5xy)<=(3/căn10)^(xy+5y^2) .

A. $\frac{5}{2}$

B. $\frac{1}{4}$

C. $\frac{1}{5}$

D. $\frac{5}{4}$

Trả lời

Đáp án đúng là: B

Ta có: ${\left(\frac{10}{9}\right)}^{2x^2-5xy}\le {\left(\frac{3}{\sqrt{10}}\right)}^{xy+5y^2}\iff {\left(\frac{10}{9}\right)}^{2x^2-5xy}\le {\left(\frac{10}{9}\right)}^{\frac{-(xy+5y^2)}{2}}$

$\iff 2x^2-5xy\le \frac{-(xy+5y^2)}{2}\iff 4x^2-9xy+5y^2\le 0$

$\iff (4x-5y)(x-y)\le 0\iff \left[\begin{array}{l}\left\{\begin{array}{l}4x-5y\le 0\\ x-y\ge 0\end{array}\right.\\ \left\{\begin{array}{l}4x-5y\ge 0\\ x-y\le 0\end{array}\right.\end{array}\right.\iff \left\{\begin{array}{l}\frac{x}{y}\le \frac{5}{4}\\ \frac{x}{y}\ge 1\end{array}\right.$

Vậy $\min \frac{x}{y}=1;\max \frac{x}{y}=\frac{5}{4}$

Vậy $\max \frac{x}{y}-\min \frac{x}{y}=\frac{5}{4}-1=\frac{1}{4}$ .