Question

先化簡(jiǎn)代數(shù)式（

x2-y2

x2-y2

-

x-y

x+y

）÷

2xy

(x-y)2(x+y)

，然后請(qǐng)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(shù)式有意義）

Answer 1

分析：本題的關(guān)鍵是正確進(jìn)行分式的通分、約分，并準(zhǔn)確代值計(jì)算．要注意最后取的數(shù)值x≠y或x≠-y．

解答：解：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．

點(diǎn)評(píng)：．取數(shù)代入求值時(shí)，要特注意原式及化簡(jiǎn)過(guò)程中的每一步都有意義，如果取的數(shù)值x=y或x=-y，則原式?jīng)]有意義．