Question

Cho x\(^2\)+y\(^2\)=\(\frac{25xy}{12}\)và x>y>0

Tính giá trị biểu thức Q=\(\frac{x+y}{x-y}\)

Answer 1

\(12\left(x^2+y^2\right)=25xy\Leftrightarrow\frac{12\left(x^2+y^2\right)}{xy}=25\)

\(\Leftrightarrow12\left(\frac{x}{y}+\frac{y}{x}\right)=25\)

Đặt \(\frac{x}{y}=t>1\Rightarrow12\left(t+\frac{1}{t}\right)=25\Leftrightarrow12t^2-25t+12=0\) \(\Rightarrow t=\frac{4}{3}\)

\(\Rightarrow Q=\frac{\frac{x}{y}+1}{\frac{x}{y}-1}=\frac{t+1}{t-1}=\frac{\frac{4}{3}+1}{\frac{4}{3}-1}=7\)