若丨x-3丨 丨y 4丨 丨z-5丨=0,求x,y,z的乘积
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/17 01:31:38
![若丨x-3丨 丨y 4丨 丨z-5丨=0,求x,y,z的乘积](/uploads/image/f/6970224-48-4.jpg?t=%E8%8B%A5%E4%B8%A8x-3%E4%B8%A8+%E4%B8%A8y+4%E4%B8%A8+%E4%B8%A8z-5%E4%B8%A8%3D0%2C%E6%B1%82x%2Cy%2Cz%E7%9A%84%E4%B9%98%E7%A7%AF)
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
按x得降幂排列:x^4-4x^3y-x²y²+3xy^3-y^4按y得升幂排列:x^4-4x^3y-x²y²+3xy^3-y^4
丨X+2丨+(Y-3)²=0x+2=0且y-3=0即:x=-2且y=3代入(x/5-2y+z)+5=y/2+x+z-2/5-6+z+5=3/2-2+z-7/5≠-1/2原式不成立再问:我是要
这个题目没有问题么,我是说最后一个式子确定是z+5y+8z=-2?如果没有问题的话:x+y+z=1;①x+3y+7z=-1;②z+5y+8z=-2③①-②2Y+6Z=-2Y=(-2-6Z)/2=-1-
根据题意,得x+y2=3x−2yx+y2=10+6x+y4,整理得x−y=0(1)4x−y=−10(2),由(1)-(2),并解得x=-103(3).把(3)代入(1),解得y=-103,所以原方程组
25xy^2z^2(x+y-z)-30xyz(z-x-y)^2+5xyz^3(z-x-y)=25xy^2z^2(x+y-z)+30xyz(x+y-z)^2-5xyz^3(x+y-z)=5xyz(x+y
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-
设x/2=y/3=z/5=ax=2ay=3az=5a是不是求的是:(x+3y-z)/(x-3y+z)?若是,如下:(x+3y-z)/(x-3y+z)=(2a+9a-5a)/(2a-9a+5a)=-3