1、已知x^2+y^2+6x-4y+13=0,则(x+y)^2006的值为_____.
问题描述:
1、已知x^2+y^2+6x-4y+13=0,则(x+y)^2006的值为_____.
2、若x^2-x+1=0,则x^2001=______.
3、若a为有理数,且x=2004a+2003,y=2004a+2004,z=2004a+2005.求x^2+y^2+z^2-xy-yz-zx的值.
4.分解因式:-4m^3+16m^2-12m
答
x^2+y^2+6x-4y+13=0(x^2+6x+9)+(y^2-4y+4)=0(x+3)^2+(y-2)^2=0x+3=0y-2=0x=-3y=2(x+y)^2006=1x^2-x+1=0(x+1)(x^2-x+1)=0x^3+1=0x^3=-1x=-1x^2001=-1x^2+y^2+z^2-xy-yz-zx=(2x^2+2y^2+2z^2-2xy-2yz-2zx)/2=[(x^2-...