Question

x>y>0 , x²+y²=6xy then x+y/x-y =

Answer 1

Given,
x²+y²=6xy.......(1)
We know that (x+y)² = x²+y²+2xy    ⇒ x²+y² = (x+y)²-2xy..............(2)
                        (x-y)² = x²+y²-2xy      ⇒ x²+y² = (x-y)²+2xy ..............(3)
From (1) , (2) and (3)
(x+y)²-2xy = 6xy
(x+y)² = 8xy
x+y = 2$\sqrt{2xy}$

(x-y)²+2xy = 6xy
(x-y)² = 4xy
x-y = 2$\sqrt{xy}$

$\frac{x+y}{x-y}$ = 2$\sqrt{2xy}$ / 2$\sqrt{xy}$
$\frac{x+y}{x-y}$ = $\frac{1}{ \sqrt{2} }$

Answer 2

Answer:

x²+y²=6xy.......(1)

We know that (x+y)² = x²+y²+2xy ⇒ x²+y² = (x+y)²-2xy..............(2)

(x-y)² = x²+y²-2xy ⇒ x²+y² = (x-y)²+2xy ..............(3)

From (1) , (2) and (3)

(x+y)²-2xy = 6xy

(x+y)² = 8xy

x+y = 2\sqrt{2xy}

2xy

(x-y)²+2xy = 6xy

(x-y)² = 4xy

x-y = 2\sqrt{xy}

xy

\frac{x+y}{x-y}

x−y

x+y

= 2\sqrt{2xy}

2xy

/ 2\sqrt{xy}

xy

\frac{x+y}{x-y}

x−y

x+y

= \frac{1}{ \sqrt{2} }

2

1