解方程组x+2y+3z=根号(x+2y+3z+2)+4

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：2024/11/10 14:08:22

解方程组x+2y+3z=根号(x+2y+3z+2)+4
x/2=y/3=z/4

令k= x/2=y/3=z/4
x=2k,y=3k,z=4k
令a=√(x+2y+3z+2)
则a²=x+2y+3z+2
x+2y+3z=a²-2
所以a²-2=a+4
a²-a-6=0
(a-3)(a+2)=0
a=√(x+2y+3z+2)>=0
所以a=3
x+2y+3z=a²-2=7
把x=2k,y=3k,z=4k代入
2k+6k+12k=7
20k=7
k=7/20
x=2k=7/10
y=3k=21/20
z=4k=7/5

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