Question

用换元法解分式方程:x-3分之2（x-1）+x-1分之6（x+3）=7

Answer 1

是6（x-3）吧

设（x-1）/（x-3）=y，原分式方程化为
2y+6/y=7
2y�0�5+6-7y=0
2y�0�5-7y+6=0
（2y-3）（y-2）=0
∴2y-3=0——y=3/2
y-2=0——y=2
∴（x-1）/（x-3）=3/2时
解得x=7
（x-1)/(x-3）=2时
解得x=5

Answer 2

解 设 a=x-3 x-1=a+2
2（a+2）/a+ 6a/(a+2) =7
2（a+2）^2+ 6a^2=7a(a+2)
a^2 - 6a+8=o
(a-6)(a-2) =0
a=6 x=9
a=2 x=5