已知方程(k^2-1)x^2+(k+1)x+(k+7)y=k-2.
1个回答
(1)k^2-1=0,且k+1=0,k+7≠0时为一元一次方程,解得:k=正负1
(2)k^2-1=0时,但k+1≠0,k+7≠0时二元一次方程,解得:k=1
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