作业帮3道方程要过程.新年快乐!
来源：学生作业帮 编辑：作业帮 分类：综合作业 时间：2024/11/08 00:01:58
3道方程要过程.新年快乐!
①
(1-x)/(x-2)-1/(2-x)=-2
(1-x)/(x-2)+1/(x-2)=-2
(1-x)/(x-2)-1/(2-x)=-2
(1-x+1)/(x-2)=-2
(2-x)/(x-2)=-2
2-x=-2(x-2)
2-x=-2x+4
2x-x=4-2
x=2
检验：当x=2时,分母为0,即x=2是增根
所以原方程无解
②
乘以(x+2)(x-2)得
(x-2)²-16=(x+2)(x-2)
x²-4x+4-16=x²-4
-4x-12=-4
-4x=8
x=-2
检验：当x=-2时,出现分母为0,即x=-2是增根
所以原方程无解
化简
原式
=(x-2)/(x-1)÷[(x+1)(x-1)-3]/(x-1)
=(x-2)/(x-1)÷[x²-1-3]/(x-1)
=(x-2)/(x-1)÷(x+2)(x-2)/(x-1)
=(x-2)/(x-1)×(x-1)/[(x+2)(x-2)]
=1/(x+2)
(1-x)/(x-2)-1/(2-x)=-2
(1-x)/(x-2)+1/(x-2)=-2
(1-x)/(x-2)-1/(2-x)=-2
(1-x+1)/(x-2)=-2
(2-x)/(x-2)=-2
2-x=-2(x-2)
2-x=-2x+4
2x-x=4-2
x=2
检验：当x=2时,分母为0,即x=2是增根
所以原方程无解
②
乘以(x+2)(x-2)得
(x-2)²-16=(x+2)(x-2)
x²-4x+4-16=x²-4
-4x-12=-4
-4x=8
x=-2
检验：当x=-2时,出现分母为0,即x=-2是增根
所以原方程无解
化简
原式
=(x-2)/(x-1)÷[(x+1)(x-1)-3]/(x-1)
=(x-2)/(x-1)÷[x²-1-3]/(x-1)
=(x-2)/(x-1)÷(x+2)(x-2)/(x-1)
=(x-2)/(x-1)×(x-1)/[(x+2)(x-2)]
=1/(x+2)