Question

Cho x>y>0 và 2x²+2y²=5xy. Tính : \(E=\frac{x+y}{x-y}\).

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=\(\frac{x+y}{x-y}\)=\(\frac{x+2x}{x-2x}\)=\(\frac{3x}{-1x}\)= -3

Answer 2

2x^2 + 2y^2 = 5xy

<=> 2x^2 + 2y^2 - 5xy = 0

<=> 2x^2  - 4xy + 2y^2 - xy  = 0

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

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

<=> 2x = y hoặc x = 2y

thay vào là xong

Answer 3

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

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

=>\(E=3\)

Answer 4

Cảm ơn olm nhiều ạ