莫非MM
幼苗
共回答了18个问题采纳率:88.9% 举报
三个方程相加:
(k+2)(x+y+z)=3
k=-2,则方程无解,所以k≠-2,k+2≠0
x+y+z=3/(k+2),
用该方程依次减去三个已知方程:
(1-k)x=3/(k+2),
x=3/[(k+2)(1-k)]
(1-k)y=3/(k+2)-1=(1-k)/(k+2),
k≠1,k-1≠0,
y=1/(k+2)
(1-k)z= 3/(k+2)- 2=(-2k-1)/(k+2),
z=(2k+1)/[(k-1)(k+2)]
所以方程组的解为:
x=3/[(k+2)(1-k)]
y=1/(k+2)
z=(2k+1)/[(k-1)(k+2)]
1年前
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