ĐK: \(y\ne0,xy\ge0\).
\(4x^2+9y^2=16xy\)
Chia cả hai vế cho \(y^2\)ta được:
\(4\left(\frac{x}{y}\right)^2+9=\frac{16x}{y}\)
\(\Leftrightarrow\frac{x}{y}=\frac{4\pm\sqrt{7}}{2}\)
Với \(y>0\)thì \(x\ge0\)
\(P=\frac{\sqrt{xy}+\sqrt{y^2}}{y}-\sqrt{\frac{x}{y}}=\frac{\sqrt{x}\sqrt{y}+y}{y}-\sqrt{\frac{x}{y}}=\sqrt{\frac{x}{y}}+1-\sqrt{\frac{x}{y}}=1\)
Với \(y< 0\)thì \(x\le0\):
\(P=\frac{\sqrt{xy}+\sqrt{y^2}}{y}-\sqrt{\frac{x}{y}}=\frac{\sqrt{-x}\sqrt{-y}-y}{y}-\sqrt{\frac{x}{y}}=-\sqrt{\frac{x}{y}}-1-\sqrt{\frac{x}{y}}=-2\sqrt{\frac{x}{y}}-1\)
\(=-2\sqrt{\frac{4\pm\sqrt{7}}{2}}-1=-\left(1\pm\sqrt{7}\right)-1=-2\pm\sqrt{7}\)
