解下列分式方程:(1)7/(x-2)=5/x;(2)x/(x-1)+(5x-2)/(x²-x)=1;
问题描述：
解下列分式方程:(1)7/(x-2)=5/x;(2)x/(x-1)+(5x-2)/(x²-x)=1;
(3)2/(x+3)+3/2=7/(2x+6);
(4)1/(x²+5x+6)=1/(x+2)+1/(x+3).
答
:(1)7/(x-2)=5/x
7x=5x-10
2x=-10
x=-5
(2)x/(x-1)+(5x-2)/(x²-x)=1;
x²+5x-2=x²-x
5x+x=2
6x=2
x=-1/3
(3)2/(x+3)+3/2=7/(2x+6);
4+3x+9=7
3x=7-13
3x=-6
x=-2
(4)1/(x²+5x+6)=1/(x+2)+1/(x+3).
1/(x+2)(x+3)=1/(x+2)+1/(x+3)
1=x+3+x+2
-2x=5-1
-2x=4
x=-2
∵x=-2原方程无意义
∴原方程无解两边同乘以多少?(1)两边同乘以 x(x-2) (2)两边同乘以 x(x-1) (3)两边同乘以 2(x+3) (4)两边同乘以 (x+2)(x+3)