Pfrac{x-y}{x+y}Rightarrow P^2frac{left(x-yright)^2}{left(x+yright)^2}Ta có: 2x^2+2y^25xyleft(1right)Leftrightarrow2x^2-4xy+2y^2xyLeftrightarrow2left(x-yright)^2xyLeftrightarrowleft(x-yright)^2frac{xy}{2}(2)Từ left(1right)Rightarrow2x^2+4xy+2y^29xyLeftrightarrow2left(x+yright)^29xyLeftrightarrowleft(x+yright)^2frac{9xy}{2}(3)Thay (2)và (3) vào P^2ta được:P^2frac{left(x-yright)^2}{left(x+yright)^2}frac{xy}{2}.frac{2}{9xy}frac{1}{9}Nhận thấy yx0Rightarrow x-y 0Rightarrowfrac{x-y}{x+y} 0Rightarrow P...
Đọc tiếp
\(P=\frac{x-y}{x+y}\)\(\Rightarrow P^2=\frac{\left(x-y\right)^2}{\left(x+y\right)^2}\)
Ta có: \(2x^2+2y^2=5xy\left(1\right)\)
\(\Leftrightarrow2x^2-4xy+2y^2=xy\)
\(\Leftrightarrow2\left(x-y\right)^2=xy\)
\(\Leftrightarrow\left(x-y\right)^2=\frac{xy}{2}\)(2)
Từ \(\left(1\right)\Rightarrow2x^2+4xy+2y^2=9xy\)
\(\Leftrightarrow2\left(x+y\right)^2=9xy\)
\(\Leftrightarrow\left(x+y\right)^2=\frac{9xy}{2}\)(3)
Thay (2)và (3) vào \(P^2\)ta được:
\(P^2=\frac{\left(x-y\right)^2}{\left(x+y\right)^2}=\frac{xy}{2}.\frac{2}{9xy}=\frac{1}{9}\)
Nhận thấy \(y>x>0\Rightarrow x-y< 0\Rightarrow\frac{x-y}{x+y}< 0\Rightarrow P< 0\)
\(\Rightarrow P=\frac{-1}{3}\)