x+y+z=0 求证:(x^2-y^2)+(xz-yz)=0

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x+y+z=0 求证:(x^2-y^2)+(xz-yz)=0

x+y+z=0 求证:(x^2-y^2)+(xz-yz)=0
x+y+z=0 求证:(x^2-y^2)+(xz-yz)=0

x+y+z=0 求证:(x^2-y^2)+(xz-yz)=0
(x^2-y^2)+(xz-yz)
=(x-y)(x+y)+z(x-y)
=(x-y)(x+y+z)
=0

原式=(x+y)(x-y)+(x-y)z=(x-y)(x+y+z)=0

证明:
(x^2-y^2)+(xz-yz)
=(x+y)(x-y)+(x-y)*z
因为x+y+z=0
所以x+y=-z,代入原式
=-z*(x-y)+(x-y)*z
=0
得证