Lời giải:
Đặt $x=ty$ ($0< t< 2$)
\(2x^2+y^2=5xy\)
\(\Leftrightarrow 2t^2y^2+y^2-5ty^2=0\)
\(\Leftrightarrow y^2(2t^2-5t+1)=0\Rightarrow 2t^2-5t+1=0\) (Do $y\neq 0$)
\(\Leftrightarrow 2(t-\frac{5}{4})^2=\frac{17}{8}\Rightarrow t-\frac{5}{4}=\pm \frac{\sqrt{17}}{4}\)
\(\Rightarrow t=\frac{5\pm \sqrt{17}}{4}\). Mà $0< t< 2$ nên $t=\frac{5-\sqrt{17}}{4}$
Do đó:
\(D=\frac{x+y}{x-y}=\frac{ty+y}{ty-y}=\frac{y(t+1)}{y(t-1)}=\frac{t+1}{t-1}=\frac{\frac{5-\sqrt{17}}{4}+1}{\frac{5-\sqrt{17}}{4}-1}=\frac{1-\sqrt{17}}{2}\)
Lời giải:
Đặt $x=ty$ ($0< t< 2$)
\(2x^2+y^2=5xy\)
\(\Leftrightarrow 2t^2y^2+y^2-5ty^2=0\)
\(\Leftrightarrow y^2(2t^2-5t+1)=0\Rightarrow 2t^2-5t+1=0\) (Do $y\neq 0$)
\(\Leftrightarrow 2(t-\frac{5}{4})^2=\frac{17}{8}\Rightarrow t-\frac{5}{4}=\pm \frac{\sqrt{17}}{4}\)
\(\Rightarrow t=\frac{5\pm \sqrt{17}}{4}\). Mà $0< t< 2$ nên $t=\frac{5-\sqrt{17}}{4}$
Do đó:
\(D=\frac{x+y}{x-y}=\frac{ty+y}{ty-y}=\frac{y(t+1)}{y(t-1)}=\frac{t+1}{t-1}=\frac{\frac{5-\sqrt{17}}{4}+1}{\frac{5-\sqrt{17}}{4}-1}=\frac{1-\sqrt{17}}{2}\)
