将(x+y+z)²展开有
(x+y+z)²=x²+y²+z²+2xy+2xz+2yz
=x²+y²+z²
所以 2xy+2xz+2yz=0
而要证明的式子=xy+xz+yz+yx+zx+zy
=2xy+2xz+2yz=0
所以命题成立
已知(x+y+z)^2=x^2+y^2+z^2,证明x(y+z)+y(z+x)+z(x+y)=0
2个回答
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