xy(x²-y²)+yz(y²-z²)+zx(z²-x²)=xy(x²-y²)+yz(y²-x²+x²-z²)+zx(z²-x²)=xy(x²-y²)-yz(x²-y²)+yz(x²-z²)-zx(x²-z²)=(...
xy(x2-y2)+yz(y2-z2)+zx(z2-x2) 因式分解 对称多项式 相关知识点: 试题来源: 解析 xy(x2-y2)+yz(y2-x2+x2-z2)+zx(z2-x2) =(xy-yz)(x2-y2)+(yz—xz)(x2-z2)=(x-y)(x-z)(y(x+y)-z(x+z))=(x-y)(x-z)(x+y+z)(y-z)...
证明：因为x^2-yz=y^2-zx=z^2-xy,所以x^2-yz=y^2-zx得x^2-y^2+zx-yz=（x+y）*（x-y）+z（x-y）=0 即x-y=0或x+y+z=0，同理y^2-zx=z^2-xy得到y-z=0或x+y+z=0
xy(x2-y2)+yz(y2-z2)+zx(z2-x2)=x^3y-xy^3+y^3z-yz^3+z^3x-zx^3 =(y-z)x^3-(x-z)y^3+(x-y)z^3 =(y-z)x^3-(x-y+y-z)y^3+(x-y)z^3 =(y-z)(x^3-y^3)+(x-y)(z^3-y^3)=(x-y)(y-z)(x^2+xy+y^2-z^2-zy-y^2)=(x-y)(y-z)(...
分解因式:xy(x2+y2)+yz(y2+z2)+zx(z2+x2)+xyz(x+y+z) 答案 (x2+y2+z2)(xy+yz+zx).由因式定理可知,原式没有一次因式,故可设原式=[a(x2+y2+z2)+b(xy+yz+zx)][c(x2+y2+z2)+d(xy+yz+zx)],比对系数可知,xy(x2+y2)+yz(y2+z2)+zx(z2+x2)+xyz(x+y+z)=(x2...
注;x^2就是x的平方 xy（x^2-y^2）+yz（y^2-z^2）+zx（z^2-x^2）解:原式=xy(x+y)(x-y)+yz(y+z)(y-z)+zx(x+z)(x-z)提取y得 =y(x+z)(x+y)(x-y)(y+z)(y-z)+zx(x+z)(x-z)提取(x+z)得 =(y+zx)(x+z)(x-z)(x+y)(x-y)(y+z)(y-z)如有不...
答案解析 查看更多优质解析 解答一 举报 xy(x2-y2)+yz(y2-x2+x2-z2)+zx(z2-x2) =(xy-yz)(x2-y2)+(yz—xz)(x2-z2)=(x-y)(x-z)(y(x+y)-z(x+z))=(x-y)(x-z)(x+y+z)(y-z) 解析看不懂?免费查看同类题视频解析查看解答 更多答案(1) ...
(列1-列2)提取(x-y),(列2-列3)提取(y-z)得| x+y y+z ||x²+xy+y² y²+yz+z²|(列1-列2)提取(x-z)得| 1 y+z ||x+y+z y²+yz+z²|展开得xy+yz+zx将所有提取的因子乘起来得原行列式=(xy+yz+zx)(x-y)(y-z)(x-z) ...
不等式两边同时乘以2，得2（x²+y²+z²）≥2（xy+yz+zx）去括号得：x²+y²+z²+x²+y²+z²≥xy+yz+zx+xy+yz+zx x²+y²≥2xy，y²+z²≥2yz，x²+z²≥2zx，代入原式化简可得x²+y&...
◎ 题干 2x-y-z x2-xy-xz+yz 2y-x-z y2-xy-yz+xz 2z-x-y z2-xz-yz+xy ◎ 答案单词记不住,就用智驭词 (点击了解) 查看答案 ◎ 解析 查看解析