Question

cho x>y>0 va 2x²+2y²=5xy

tinh gia tri bieu thuc E=x+y/x-y

Answer 1

2x²+2y²=5xy

<=>2x²-5xy+2y²=0

<=>(2x²-4xy)-(xy-2y²)=0

<=>2x(x-2y)-y(x-2y)=0

<=>(x-2y)(2x-y)=0

<=> x-2y=0 hoặc 2x-y=0

*)Nếu x-2y=0=>x=2y

=>E=\(\frac{x+y}{x-y}=\frac{2y+y}{2y-y}=\frac{3y}{y}=3\)

*)Nếu 2x-y=0=>2x=y

=>E=\(\frac{x+y}{x-y}=\frac{x+2x}{x-2x}=\frac{3x}{-x}=-3\)

Answer 2

Ta có: x>y>0

\(\Rightarrow\hept{\begin{cases}x+y>0\\x-y>0\end{cases}}\)

\(\Rightarrow E=\frac{x+y}{x-y}>0\)

Ta có : E\(=\frac{x+y}{x-y}\)

\(\Rightarrow E^2=\frac{\left(x+y\right)^2}{\left(x-y\right)^2}=\frac{x^2+2xy+y^2}{x^2-2xy+y^2}=\frac{2\left(x^2+2xy+y^2\right)}{2\left(x^2-2xy+y^2\right)}=\frac{2x^2+4xy+2y^2}{2x^2-4xy+2y^2}\)\(=\frac{5xy+4xy}{5xy-4xy}=\frac{9xy}{xy}=9\)

\(\Rightarrow E=\sqrt{9}\)( do E>0)

\(\Leftrightarrow E=3\)