得到两组解
0,0,0
0.2119268995,0.2119268995,0.2119268995
解方程组4x^2/(1+4x^2)=y,4y^2/(1+4y^2)=z,4z^2/(1+4z^2)=x
2个回答
相关问题
-
解三元分式方程组 y= (4x^2)/(1+4x^2 ) z= (4y^2)/(1+4y^2 ) x= (4z^2)/(
-
解方程组1/x＋2/y+3/z=-1,2/x-4/y-9/z=-1,4/x+6/y-9/z=2
-
解方程组{X-2Y+Z=0 {2X+Y-Z=1 {3Z+2Y-Z=4
-
解方程组[ 5x+y+z=1 ,2x-y+2z=1,x +5y-z=-4
-
解方程组(1){3x+2y+z=7,2x+3y+4z=3,x+4y+6z=1
-
解方程组1.2x+y+z=7 2.x+2y+z=5 3.x+y+2z=4
-
解方程组14x-4y+2z=9 3x+2y+4z=-2 6x-4y-2z=7
-
解方程组{3x+2y=1,2x-y+2z=-4,x+2y-z=3.
-
解方程组：y+2x=-2,3x+y-4z=1,2x-y+2z=8
-
x^2+y^2+z^2+4x+4y+4z+1=0,求x+y+z