本团长来帮你.
①已知x=2009,y=2010,求代数式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先化简,再求值：（a^2-b^2/a^2-ab）÷（a+（2ab+b^2/a)当b=-1时,再从-2＜a＜2的范围内选取一个合适的整数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 设a=0
=1/(0-1)
=-1