解下列分式方程(组): (1)
(2)
|
(1)原式可化为
(5-
)+(1+1 x-19
)=(4+1 x-9
)+(2+5 x-6
),5 x-8
即
-1 x-9
=1 x-19
+5 x-6
,5 x-8
∴
=-10 (x-9)(x-19)
,-10 (x-6)(x-8)
∴(x-6)(x-8)=(x-9)(x-19),
即14x=123,
∴x=
,123 14
经检验x=
是原方程的解,123 14
故x=
;123 14
(2)原方程可化为
+1 a
=3①,1 b
+1 b
=4②,1 c
+1 c
=5③,1 a
①+②+③得
+1 a
+1 b
=6④,1 c
④-①得
c=
,1 3
④-②得
a=
,1 2
④-③得
b=1,
经检验a=
,b=1,c=1 2
是原方程的解,1 3
∴
.a= 1 2 b=1 c= 1 3