本團(tuán)長(zhǎng)來幫你.
①已知x=2009,y=2010,求代數(shù)式x-y/x÷（x-（2xy-y^2/x))的值
(x-y)/x÷[x-（2xy-y^2/x)]
=(x-y)/x÷(x²-2x+y²)/x
=(x-y)/x÷(x-y)²/x
=1/(x-y)
=1/(2009-2010)
=-1
②先化簡(jiǎn),再求值：（a^2-b^2/a^2-ab）÷（a+（2ab+b^2/a)當(dāng)b=-1時(shí),再?gòu)?2＜a＜2的范圍內(nèi)選取一個(gè)合適的整數(shù)a代入求值
（a²-b²)/(a²-ab）÷[a+（2ab+b²/a)]
=[(a-b)(a+b)/a(a-b)]÷(a²+2ab+b²)/a
=(a+b)/a÷(a+b)²/a
=1/(a+b) b=-1 設(shè)a=0
=1/(0-1)
=-1