已知1999(x-y)+2000(y-z)-2001(x-y)=0
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/10/02 20:35:22
已知1999(x-y)+2000(y-z)-2001(x-y)=0
1999²(x-y)+2000²(y-z)+2001²(y-z)=2000,求z-y的值
1式的2001(x-y)应改为2001(x-z),二式的2001²(y-z)应改为2001²(z-x)
1999²(x-y)+2000²(y-z)+2001²(y-z)=2000,求z-y的值
1式的2001(x-y)应改为2001(x-z),二式的2001²(y-z)应改为2001²(z-x)
已知1999(x-y)+2000(y-z)+2001(x-z)=0
∴1999²(x-y)+2000²(y-z)+2001²(z-x)=2000,求z-y的值
1999(x-y)+2000(y-z)-2001(x-z)=0
1999x-1999y+2000y-2000z-2001x+2001z=0
-2x+y+z=0
∴x=(y+z)/2
1999²(x-y)+2000²(y-z)+2001²(z-x)=2000
1999²[(y+z)/2-y]+2000²(y-z)+2001²[(z-(y+z)/2]=2000
1999²(z-y)/2+2000²(y-z)+2001²(z-y)/2=2000
(z-y)(1999²-2*2000²+2001²)=2000
(z-y)[1999²-2*(1999+1)²+2001²]=2000
(z-y)(1999²-2*1999²-4*1999-2+2001²)=2000
(z-y)(2001²-1999²-4*1999-2)=2000
(z-y)(8000-7996-2)=2000
(z-y)*2=2000
∴z-y=1/1000
∴1999²(x-y)+2000²(y-z)+2001²(z-x)=2000,求z-y的值
1999(x-y)+2000(y-z)-2001(x-z)=0
1999x-1999y+2000y-2000z-2001x+2001z=0
-2x+y+z=0
∴x=(y+z)/2
1999²(x-y)+2000²(y-z)+2001²(z-x)=2000
1999²[(y+z)/2-y]+2000²(y-z)+2001²[(z-(y+z)/2]=2000
1999²(z-y)/2+2000²(y-z)+2001²(z-y)/2=2000
(z-y)(1999²-2*2000²+2001²)=2000
(z-y)[1999²-2*(1999+1)²+2001²]=2000
(z-y)(1999²-2*1999²-4*1999-2+2001²)=2000
(z-y)(2001²-1999²-4*1999-2)=2000
(z-y)(8000-7996-2)=2000
(z-y)*2=2000
∴z-y=1/1000
已知x+y-z/z=x-y+z/y=-x+y+z/x,且xyz不等于0,求分式[(x+y)(x+z)(y+z)]/xyz
已知:(x+y-z)/z=(x-y+z)/y+(y+z-x)/x,且xyz≠0,求代数式[(x+y)(y+z)(x+z)
已知:1998(x-y)+1999(y-z)+2000(z-x)=0
3.已知:1998(x-y)+1999(y-z)+2000(z-x)=0,1998^2(x-y)+1999^2(y-z)
已知x、y、z满足方程组:x+y-z=6;y+z-x=2;z+x-y=0 求x、y、z的值
已知(x+y+z)^2=x^2+y^2+z^2,证明x(y+z)+y(z+x)+z(x+y)=0
已知x,y,z 大于0,x+y+z=2,求证 xz/y(y+z)+zy/x(x+y)+yx/z(z+x)大于等于2/3
1998(z-y)+1999(y-z)+2000(z-x)=0 1998²(z-y)+1999²(y
已知:(x+y)/z=(x+z)/y=(z+y)/x,且xyz不等于0,则分式(x+y)(x+z)(z+x)/xyz的值
已知3x-2y-5z=0,2x-5y+4z=0,且x,y,z均不为0,求3x*x+2y*y+5z*z/5x*x+y*y-
已知x,y,z满足方程组{x+y-z=6{y+z-x=2{z+x_y=0,求x,y,z的值
已知 x/(y+z)+y/(z+x)+z/(x+y)=1