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)...
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...
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)=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)(...
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百度试题 结果1 题目计算下列第二类曲线积分:(x2-yz)dx+(y2-zx)dy+(z2-xy)dz,其中为螺旋线x=cosp,y=sin,z=φ上由点(1,0,0)到(1,0,2元)的一段弧; 相关知识点: 试题来源: 解析 83 反馈 收藏
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) ...
求采纳哦!=27 下面设 x-y=a；z-x=b；则 z-y=a+b 所以有 a^2+b^2+(a+b)^2=54 又有 a^2+b^2>=(a+b)^2/2 【那几个基本不等式】所以有 3/2*(a+b)^2<=54 所以有 |a+b|<=6 即|y-z|<=6 就求出来了 最大值是 6 ...
已知x^2-yz=y^2-zx=z^2-xy,求证x=y=z或x+y+z=0 相关知识点: 试题来源: 解析 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...
答案—2 一解析 由x2y=y2-xx x^2-y^2=z(y-x)⇒(x-y)(x+y)=z(y-x)0 同理:由)(y+z)=x(z—y)2,当y时,两 y^2-zx=z^2-xy⇒(y-z) x^2-yz=z^2-xy⇒(x-z) 由xx+2)=2y(—x)2同除从2y得 若X=y.则由2得:y+=-x=-2x y+2=—x 若x=y=,则y2-xy...