Question

先化簡代數式（

x2-y2

x2-y2

-

x-y

x+y

）÷

2xy

(x-y)2(x+y)

，然后請你任意選取一組x、y的值代入求值．（所取的x、y值要保證原代數式有意義）

Answer 1

原式=（

x2-y2

(x+y)(x-y)

-

(x-y)2

(x-y)(x+y)

）÷

2xy

(x-y)2(x+y)

=

x2-y2-(x-y)2

(x+y)(x-y)

÷

2xy

(x-y)2(x+y)

=

2xy

(x-y)(x+y)

×

(x-y)2(x+y)

2xy

=x-y，
選取x=2，y=1，∴原式=2-1=1．