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解:(k-1)/(x²-1)+1/(x-x²)=(5-k)/(x²+x)
(k-1)/[(x+1)(x-1)]-1/[x(x-1)]=(5-k)/[x(x+1)] 方程两边同时乘x(x+1)(x-1)
(k-1)x-(x+1)=(5-k)(x-1)
(k-1)x-x-1=(5-k)x-5+k
(k-1)x-x-(5-k)x=-5+k+1
(k-1-1-5+k)x=k-4
(2k-7)x=k-4
x=(k-4)/(2k-7)
当x(x+1)(x-1)=0时,方程有增根,这时:x=0或x=-1或x=1
当x=0时,有:k-4=0, k=4;
当x=-1时,有:(k-4)/(2k-7)=-1
k-4=7-2k
3k=11
k=11/3
当x=1时,有:(k-4)/(2k-7)=1
k-4=2k-7
k=3
所以,原分式方程有增根时,k=4或k=11/3或k=3
(k-1)/[(x+1)(x-1)]-1/[x(x-1)]=(5-k)/[x(x+1)] 方程两边同时乘x(x+1)(x-1)
(k-1)x-(x+1)=(5-k)(x-1)
(k-1)x-x-1=(5-k)x-5+k
(k-1)x-x-(5-k)x=-5+k+1
(k-1-1-5+k)x=k-4
(2k-7)x=k-4
x=(k-4)/(2k-7)
当x(x+1)(x-1)=0时,方程有增根,这时:x=0或x=-1或x=1
当x=0时,有:k-4=0, k=4;
当x=-1时,有:(k-4)/(2k-7)=-1
k-4=7-2k
3k=11
k=11/3
当x=1时,有:(k-4)/(2k-7)=1
k-4=2k-7
k=3
所以,原分式方程有增根时,k=4或k=11/3或k=3
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