Đáp án: $x=y=9$
Giải thích các bước giải:
ĐKXĐ: $x>0,y>0$
Ta có:
$\dfrac{x^2+y^2}{xy}+\dfrac{2\sqrt{xy}}{x+y}=3$
$\to 2+\dfrac{x^2+y^2}{xy}+\dfrac{2\sqrt{xy}}{x+y}=5$
$\to\dfrac{x^2+y^2+2xy}{xy}+\dfrac{2\sqrt{xy}}{x+y}=5$
$\to\dfrac{(x+y)^2}{xy}+\dfrac{2\sqrt{xy}}{x+y}=5$
Đặt $\dfrac{\sqrt{xy}}{x+y}=t, t>0$
Ta có: $\dfrac{\sqrt{xy}}{x+y}\le\dfrac{\sqrt{xy}}{2\sqrt{xy}}=\dfrac12$
$\to \dfrac{1}{t^2}+2t=5$
$\to 1+2t^3=5t^2$
$\to t\in\{\dfrac12, 1\pm\sqrt2\}$
Vì $t>0, t\le\dfrac12\to t=\dfrac12$
$\to \dfrac{\sqrt{xy}}{x+y}=\dfrac12$
Do $\dfrac{\sqrt{xy}}{x+y}\le \dfrac12$
Dấu = xảy ra khi $x=y$
Ta có $6\sqrt{x}=y+9$
$\to 6\sqrt{x}=x+9$
$\to x-6\sqrt{x}+9=0$
$\to (\sqrt{x}-3)^2=0$
$\to \sqrt{x}=3$
$\to x=9$
$\to y=x=9$
