若丨x-3丨 丨y 4丨 丨z-5丨=0,求x,y,z的乘积
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1.x+y+z2.-x-3y-1
1x^3-3x+2=(x^3-1)-3x+3=(x-1)(x^2+x+1)-3(x-1)=(x-1)(x^2+x-2)=(x-1)(x-1)(x+2)23x^3+7x^2-4=(3x^3+3x^2)+
把原式两边对x求导得:x^2+12y^3*dy/dx+1+2dy/dx=0合并同类项移项得:dy/dx=-(1+2x)/(12y^3+2)
x^2-4(x-1)=x^2-4x+4=(x-2)^2x^4-y^4=(x^2+y^2)(x^2-y^2)=(x^2+y^2)(x+y)(x-y)(5a^2+2a)-4(2+2a^2)=5a^2+2a
原式=(x4-xy3)+(y4-x3y)+(3xy2-3x2y)=x(x3-y3)+y(y3-x3)+3xy(y-x)=(x3-y3)(x-y)-3xy(x-y)=(x-y)(x3-y3-3xy)=(
解(15x^4y^4-9x^5y³-3x^6y²)/(-3x²y)²=(15x^4y^4-9x^5y³-3x^6y²)/(3x²y
由(2)得4x=3y=6z,∴x=34y,z=12y;代入(1)得:y=4,代入(2)得:x=3,z=2,方程组的解为x=3y=4z=2.
按字母x的升幂排列就把y看成系数y4-xy3+x2y2+3x3y
(x+y+z)²-(x²+y²+z²)=2(xy+yz+zx)=-1,xy+yz+zx=-1/2x3+y3+z3=3xyz+(x+y+z)(x²+y&
(x+y+z)^2=[(x+y)+z]^2=(x^2+2xy+y^2)+z^2+2zx+2zy=x^2+y^2+z^2+2xy+2xz+2yz=x^2+y^2+z^2+2(xy+xz+yz)=0x+y
原式=x4+x3y+4x3y+x2y+4x2y2+4x2y2+xy2+4xy3+xy3+y4,=x3(x+y)+4x2y(x+y)+xy(x+y)+4xy2(x+y)+y3(x+y),=-x3-4x2
因为最终结果是0,所以x-4=0,xy-z=0,所以x=4,y-z=-4,3y-3z=-12,5x=20,所以5x3y-3z=-1220=8
(x2+z2)(x2+y2)(y2+z2)=(x+y)2-2xy×(x+z)2-2xz×(y+z)2-2yz--之后不清楚了
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x4-xy3-x3y-3x2y+3xy2+y4=(x4-xy3)+(y4-x3y)+(3xy2-3x2y)=x(x3-y3)+y(y3-x3)+3xy(y-x)=(x3-y3)(x-y)-3xy(x-
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