问几道关于分式方程的数学题已知1/(x-1)=2,则2/(x-1)+x-1的值是多少?若方程1/(x+3)=2/(x+k)有负数解,则k的取值范围是?若(x-3)/(x-2)与(2-x)/(3x+2)互为倒数,则x的值是多少?已知方程[2(x+a)]/[a(x-1)]=-8/5的解为

Question

问几道关于分式方程的数学题

已知1/(x-1)=2,则2/(x-1)+x-1的值是多少?

若方程1/(x+3)=2/(x+k)有负数解,则k的取值范围是?

若(x-3)/(x-2)与(2-x)/(3x+2)互为倒数,则x的值是多少?

已知方程[2(x+a)]/[a(x-1)]=-8/5的解为x=-1/5,则a的值是多少?

Answer 1

解
∵1/(x-1)=2
∴x-1=1/2
2/(x-1)+x-1
=2×1/(x-1)+(x-1)
=2×2+1/2
=4+1/2
=9/2
1/(x+3)=2/(x+k)有负数解
两边同时乘以(x+3)(x+k)
∴x+k=2(x+3)
即x+k=2x+6
∴x=k-6<0
∴k<6
(x-3)/(x-2)与(2-x)/(3x+2)互为倒数
∴(x-3)/(x-2)×(2-x)/(3x+2)=1
∴-(x-3)/(2-x)×(2-x)/(3x+2)=1
∴-(x-3)/(3x+2)=1
∴3-x=3x+2
∴4x=1
∴x=1/4
[2(x+a)]/[a(x-1)]=-8/5
-8a(x-1)=10(x+a)
-8ax+8a=10x+10a
∴-8a×(-1/5)+8a=-2+10a
∴8a/5=-2+2a
∴8a=-10+10a
∴10=2a
∴a=5