(x+y+z)^2-(x-y-z)^2的解同上
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(x+y+z)^2-(x-y-z)^2的解同上
(x+y+z)^2-(x-y-z)^2的解
同上
(x+y+z)^2-(x-y-z)^2的解同上
使用平方差公式
(X+Y+Z)²-(X-Y-Z)²
=(X+Y+Z+X-Y-Z)〔X+Y+Z-(X-Y-Z)〕
=(X+Y+Z+X-Y-Z)(X+Y+Z-X+Y+Z)
=2X×(2Y+2Z)
=4XY+4XZ
(x+y+z)^2-(x-y-z)^2=(x+y+z+x-y-z)(x+y+z-x+y+z)=2x(2y+2z)=4xy+4xz
这个就是平方差公式
2x*2(y+z)=4xy+4xz
用行列式的性质证明:y+z z+x x+y x y z x+y y+z z+x =2 z x y z+x x+y y+z y z x 这个怎么证?
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