UC知道两道初中数学题...
来源：百度知道 编辑：UC知道 时间：2024/05/14 06:43:25
1.解方程
1/(x-1)-1/(x-2)=1/(x-3)-1/(x-4) [不能死算 还要过程]
2.若分式方程(k-1)/(x^2-1)-1/(x^2-x)=(k-5)/(x^2+x)有增根x=-1,那么k的值为( )
1/(x-1)-1/(x-2)=1/(x-3)-1/(x-4) [不能死算 还要过程]
2.若分式方程(k-1)/(x^2-1)-1/(x^2-x)=(k-5)/(x^2+x)有增根x=-1,那么k的值为( )
1/(x-1)-1/(x-2)=1/(x-3)-1/(x-4)
1/[(x-1)(x-2)]=1/[(x-3)(x-4)]
(x-1)(x-2)=(x-3)(x-4)
x^2-3x+2=x^2-7x+12
x=5/2
2.(k-1)/(x^2-1)-1/(x^2-x)=(k-5)/(x^2+x)
去分母整理得：x=(6-k)/2=1,k=4
1/(x-1)-(x-2)=1/(x-3)-1(x-4) x-2-x+1/(x-1)(x-2)=x-4-x+3/(x-3)(x-4) -1/xXx-3x+2=-1/xXx-7x+12 4x=10x x=2/5
1、1/(x-1)-1/(x-2)=1/(x-3)-1/(x-4)
1/[(x-1)(x-2)]=1/[(x-3)(x-4)]
两边同时乘以(x-1)(x-2)(x-3)(x-4)得
(x-3)(x-4)=(x-1)(x-2)
x^2-7x+12=x^2-3x+2
4x=10
x=2.5
2、先不考虑增根，将方程化简
方程两边同时乘以x(x-1)(x+1)得
(k-1)x-(x+1)=(k-5)(x-1)
kx-2x-1=kx-5x-k+5
k=6-3x
将增根x=-1带入上式得
k=9
所以k的值为9