已知x,y,z为实数,且x+y+z=1,x^2+y^2+z^2=3,则xyz的最大值是_百度作业帮
已知x,y,z为实数,且x+y+z=1,x^2+y^2+z^2=3,则xyz的最大值是
由（x+y+z）²=1,x²+y²+z²=3,x²+y²+z²+2xy+2xz+2yz=1∴xy+xz+yz=-1∵x+y+z=1,∴x+y=1-z,①代入xy+z（x+y）=-1,xy+z（1-z）=-1∴xy=-1-z+z² ②由韦达定理：Δ=（1-z）²-4（-1-z+z²）≥0,1-2z+z²+4+4z-4z²≥0-3z²+2z+5≥0,3z²-2z-5≤0,（3z-5）（z+1)\≤0-1≤z≤5/3M=xyz=（-1-z+z²）z=z³-z³-z将z=5/3代入：最大值Mmax=5/27将z=-1代入：最小值Mmin=-1.
-1x、y、z分别为-1,1,1
x+y=1-zx^2+y^2+z^2=3
x+y+z=1平方作差得xy+xz+yz=-1xy+z(x+y)=-1xy+z(1-z)=-1
xy=-1-z(1-z)
x+y=1-z看成方程判别式>=0
z在[-1,5/3]上xyz=z*(-1-z(1-z)=z^3-z^2-z求导得3z^2-2z-1导数>0有z在[-根3,-1/3]上增 在[-1/3,1]上减 在[1,根3]上增最后求得xyz最小值5/27
将x+y+z=1两边同时平方展开，得
x²+y²+z²+2(xy+yz+xz)=1
x²+y²+z²=3， 则 xy+yz+xz=-1
即 xy=-1-(x+y)z
x+y+z=1，得
xy=-1-z(1-z)=z²-z-...因式分解, 要有详细过程喔. (yz^5+zx^5+xy^5)-(zy^5+xz^5+yx^5)-(y^2*z^4+z^2*x^4+x^2*y^4)+(y^4*z^2+z^4*x^2+x^4*y^2)+xyx(yz^2+zx^2+xy^2)-xyz(zy^2+yx^2+xz^2)_百度作业帮
因式分解, 要有详细过程喔. (yz^5+zx^5+xy^5)-(zy^5+xz^5+yx^5)-(y^2*z^4+z^2*x^4+x^2*y^4)+(y^4*z^2+z^4*x^2+x^4*y^2)+xyx(yz^2+zx^2+xy^2)-xyz(zy^2+yx^2+xz^2)
(yz^5+zx^5+xy^5)-(zy^5+xz^5+yx^5)-(y^2*z^4+z^2*x^4+x^2*y^4)+(y^4*z^2+z^4*x^2+x^4*y^2)+xyx(yz^2+zx^2+xy^2)-xyz(zy^2+yx^2+xz^2)
解 分三步分解 [1],(yz^5+zx^5+xy^5)-(zy^5+xz^5+yx^5) =-yz(y^4-z^4)+zx(x^4-z^4)-xy(x^4-y^4) =-yz(y-z)(y+z)*(y^2+z^2)+zx(x-y+y-z)(z+x)(z^2+x^2)-xy(x-y)(x+y)(x^2+y^2) =-x(x-y)(y-z)[(y+z)(y^2+z^2)+x(y^2+z^2+yz)+x^2*(y+z)+x^3]+z(y-z)(x-y)[((x+y)(x^2+y^2)+z(x^2+y^2+xy)+z^2*(x+y)+z^3] =(y-z)(x-y)(z-x)[(z^3+x^3+zx(z+x)+y(z^2+x^2+zx)+zx(z+x)+y^2*(z+x)+xyz+y^3-xyz+xyz-zx(z+x)] =(y-z)(x-y)(z-x)[x^3+y^3+z^3+yz(y+z)+zx(z+x)+xy(x+y)+xyz] [2],(y^2*z^4+z^2*x^4+x^2*y^4)-(y^4*z^2+z^4*x^2+x^4*y^2) =-y^2*z^2(y-z)(y+z)+z^2*x^2(z+x)(x-y+y-z)-x^2*y^2(x-y)(x+y) =z^2*(y-z)*(x-y)[x^2+xy+y^2+z(x+y)]-x^2*(y-z)(x-y)[z^2+yz+y^2+x(y+z)] =(y-z)(x-y)(z-x)[xyz+y^2(z+x)+zx(z+x)+y(z^2+x^2+zx)] =(y-z)(x-y)(z-x)[2xyz+yz(y+z)+zx(z+x)+xy(x+y)] [3],xyx(yz^2+zx^2+xy^2)-xyz(zy^2+yx^2+xz^2) =xyz[-yz(y-z)+zx(x-y+y-z)-xy(x-y)] =xyz[z(y-z)(x-y)-x(x-y)(y-z)] =xyz(y-z)(x-y)(z-x) 所以上式分解为 (y-z)(x-y)(z-x)[x^3+y^3+z^3+yz(y+z)+zx(z+x)+xy(x+y)+xyz-2xyz-yz(y+z)-zx(z+x)-xy(x+y)+xyz] =(y-z)(x-y)(z-x)(x^3+y^3+z^3)
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您可能关注的推广回答者：回答者：已知x,y,z,a,b,均为非零实数,且满足xy/(x+y)=1/(a^3-b^3),yz/(y+z)=1/(a^3),xz/(x+z)=1/(a^3+b^3),xyz/(xy+yz+zx)=1/12,求a的值._百度作业帮
已知x,y,z,a,b,均为非零实数,且满足xy/(x+y)=1/(a^3-b^3),yz/(y+z)=1/(a^3),xz/(x+z)=1/(a^3+b^3),xyz/(xy+yz+zx)=1/12,求a的值.
前三式取倒数,有1/x+1/y=a^3-b^31/y+1/z=a^31/x+1/z=a^3+b^3叠加,得到2/x+2/y+2/z=3a^3=2(xy+yz+xz)/xyz=24a=2已知x+y+z=2 xyz=10 yz+zx+xy=-13 求yz+zx分之一+zx+zy分之一+xy+zz分之一=速速回答了..zz指的是z的平方_百度作业帮
已知x+y+z=2 xyz=10 yz+zx+xy=-13 求yz+zx分之一+zx+zy分之一+xy+zz分之一=速速回答了..zz指的是z的平方
速速回答了..zz指的是z的平方
yz+zx+xy=-13得yz ^2+z ^2x+xy z =-13 z （x+y）z^2+xy z=-13 z∴（2- z）z^2+10=-13 z z ^3-2 z ^2-13 z-10=0 得(z+1)(z+2)(z-5)=0 即z=-1,z=-2,z=5当Z=-1时原式=（yz+zx+xy）+1/z（1/x+1/y+1/z）=-13+（-1）×（-13/10）=-11.7当Z=-2时原式=（yz+zx+xy）+1/z（1/x+1/y+1/z）=-13+（-2）×（-13/10）=-12.35当Z=5时原式=（yz+zx+xy）+1/z（1/x+1/y+1/z）=-13+5×（-13/10）=-13.26
yz+zx+xy=-13得yz ^2+z ^2x+xy z =-13 z
（x+y）z^2+xy z=-13 z∴（2- z）z^2+10=-13 z
z ^3-2 z ^2-13 z-10=0
(z+1)(z+2)(z-5)=0
Z=-1或-2或5由xyz=10
yz+zx+xy=-13得1/ x +1/y+1/z=-13/10当Z=-1时...}