x+y+z=0，x的平方十y的平方十z的平方二4求x的4次方+y的4次方十z的4次方 x平方+y平方+z平方-4z=0，求z先对x后对y的二阶混合...
x\u5341y\u5341z\u4e8c0,x\u7684\u5e73\u65b9\u52a0y\u7684\u5e73\u65b9\u5341Z\u7684\u5e73\u65b9\u4e8c4\u6c42x\u76844\u6b21\u65b9\u5341y\u76844\u6b21\u65b9\u5341Z\u76844\u6b21\u65b9\u7b54\uff1a
x^4+x²y²-2y^4
=(x^4-y^4)+x²y²-y^4
=(x²+y²)(x²-y²)+(x²-y²)y²
=(x²+2y²)(x-y)(x+y)
\u91c7\u7eb3\u54e6\u2026\u2026\u2026\u2026
\u4e24\u8fb9\u5bf9x\u6c42\u5bfc\u5f97
2x+(2z-4)z'x=0
\u4e24\u8fb9\u518d\u5bf9\u6c42\u5f97\u5f97
2z'y*z'x+(2z-4)z''xy=0
所以（x+y+z)^2=0
x^2+y^2+z^2+2(xy+yz+xz)=0 (2）
因为x^2+y^2+z^2=4
所以xy+yz+xz=--2 (3)
(xy+yz+xz)^2=4
x^2y+y^2z^2+x^2z^2+2xy^2z+2x^2yz+2xyz^2=4
x^2y^2+y^2z^2+x^2z^2+2xyz(x+y+z)=4 (4)
将（1）代入（4）
x^2y^2+y^2z^2+x^2z^2=4 (5)
因为(x^2+y^2+z^2)^2=x^4+y4+z^4+2(x^2y^2+y^2z^2+x^2z^2)
x^4+y^4+z^4+2(x^2y^2+y^2z^2+x^2z^2)=16 (6)
将（5）代入（6）得
x^4+y^4+z^4=8
所以x^4+y^4+z^4的值是8
绛旓細(1)脳y-(2)寰楋細y-x+(y-z)z'=0 瑙ｅ緱锛歾'=(x-y)/(y-z)
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绛旓細瑙ｏ細鍥犱负x+y+z=0 (1)鎵浠ワ紙x+y+z)^2=0 x^2+y^2+z^2+2(xy+yz+xz)=0 (2锛夊洜涓簒^2+y^2+z^2=4 鎵浠y+yz+xz=--2 (3)(xy+yz+xz)^2=4 x^2y+y^2z^2+x^2z^2+2xy^2z+2x^2yz+2xyz^2=4 x^2y^2+y^2z^2+x^2z^2+2xyz(x+y+z)=4 (4)灏嗭紙1锛変唬鍏...
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