Question

Cho x,y thỏa mãn : \(\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)=3\sqrt{y}\left(\sqrt{x}+5\sqrt{y}\right)\)

Tính giá trị của : \(C=\frac{2x+\sqrt{xy}+3y}{x+\sqrt{xy}-y}\)

Answer 1

ĐKXĐ : x;y > 0

\(\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)=3\sqrt{y}\left(\sqrt{x}+5\sqrt{y}\right)\)

\(\Leftrightarrow x+\sqrt{xy}=3\sqrt{xy}+15y\)

\(\Leftrightarrow x=2\sqrt{xy}+15y\)

\(\Leftrightarrow\left(x-2\sqrt{xy}+y\right)-16y=0\)

\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)^2-\left(4\sqrt{y}\right)^2=0\)

\(\Leftrightarrow\left(\sqrt{x}-5\sqrt{y}\right)\left(\sqrt{x}+3\sqrt{y}\right)=0\)

Mà theo đk x;y > 0 nên \(\sqrt{x}+3\sqrt{y}>0\) Do đó \(\sqrt{x}-5\sqrt{y}=0\Rightarrow\sqrt{x}=5\sqrt{y}\Rightarrow x=25y\)

Thay vào C ta được :

\(C=\frac{2.25y+\sqrt{25y.y}+3y}{25y+\sqrt{25y.y}-y}=\frac{50y+5y+3y}{25y+5y-y}=2\)