已知x=-2010,y=2009,z=-2008,求x^2+y^2 +z^2+xy+yz-xz的值

网友回答

x²+y²+z²+xy+yz-xz
=(2x²+2y²+2z²+2xy+2yz-2xz)÷2
=[(x+y)²+(y+z)²+(x-z)²]÷2
=[(-2010+2009)²+(2009-2008)²+(-2010+2008)²]÷2
=(1+1+4)÷2
=3======以下答案可供参考======
供参考答案1:
=（2x^2+2y^2 +2z^2+2xy+2yz-2xz）/2
=[(x+y)²+(y+z)²+(x-z)²]/2
=﹙1+1+4﹚/2
=3供参考答案2:
原式=(X+Y)^2+(Y+Z)^2-2XY-2YZ-Y^2+XY+YZ-XZ=(X+Y)^2+(Y+Z)^2-XY-Y^2-YZ-XZ=(X+Y)^2+ (Y+Z)^2-Y(X+Y)-Z(X+Y)=(X+Y)^2+(Y+Z)^2-(X+Y)(Y+Z)然后把数字带入即可