[tex]\rightarrow \sqrt{x}+\sqrt{y}=\frac{1}{2}\\y=\frac{1}{36x}\\\rightarrow \sqrt{x}+\sqrt{\frac{1}{36x}}=\frac{1}{2}\\\rightarrow x+\frac{1}{36x}+\frac{1}{3}=\frac{1}{4}\\\rightarrow x+\frac{1}{36x}+\frac{1}{12}=0\\\rightarrow \frac{36x^2+1+3x}{36x}=0\\\rightarrow 36x^2+1+3x=0\\\rightarrow (3x+\frac{1}{2})^2+27x^2+\frac{3}{4}>0\\\rightarrow PT.vo.nghiem[/tex]