(1)已知(1／x) -(1／y)=3,则代数式(2x-14xy-2y)／(x-2xy-y)的值为_____.
问题描述：
(1)已知(1／x) -(1／y)=3,则代数式(2x-14xy-2y)／(x-2xy-y)的值为_____.
(2)若1／x(x+1)=(A／x)+(B／x+1),则A=_______,B=_______.
(3)分解因式：(2x)／(x^-4)+3／(x+2)+1／(2-x)；
答
1/x-1/y=3(y-x)/(xy)=3y-x=3xyx-y=-3xy(2x-14xy-2y)/(x-2xy-y)=[2(x-y)-14xy]/[(x-y)-2xy]=(-6xy-14xy)/(-3xy-2xy)=-20xy/(-5xy)=4A/x+B/(x+1)=[A(x+1)+Bx]/x(x+1)=[(A+B)x+A]/x(x+1)=1/x(x+1)所以A+B=0,A=1A=1,B=-1...