若实数x,y,z满足方程组:xyx+2y=1…(1)yzy+2z=2…(2)zxz+2x=3…(3),则有( ) A.x+2y+3z=0 B.7x+5y+2z=0 C.9x+6y+3z=0 D.10x+7y+z=0
问题描述:
若实数x,y,z满足方程组:
,则有( )
=1…(1)xy x+2y
=2…(2)yz y+2z
=3…(3)zx z+2x
A. x+2y+3z=0
B. 7x+5y+2z=0
C. 9x+6y+3z=0
D. 10x+7y+z=0
答
由(1)、(3)得y=
,z=x x−2
,6x x−3
故x≠0,代入(2)解得x=
,27 10
所以y=
,z=-54.27 7
检验知此组解满足原方程组.
∴10x+7y+z=0.
故选D.