(1)解方程：1/(x-7)+1/(x-4)=1/(x-6)+1/(x-5)
1/(x-7)+1/(x-4)=1/(x-6)+1/(x-5)
通分
(2x-11)/(x-7)(x-4)=(2x-11)/(x-6)(x-5)
(2x-11)[1/(x^2-11x+28)-1/(x^2-11x+30)]=0
因?yàn)閤^2-11x+28不等于x^2-11x+30
所以1/(x^2-11x+28)-1/(x^2-11x+30)不等于0
所以所以2x-11=0
x=11/2
分式方程要檢驗(yàn)
經(jīng)檢驗(yàn),x=11/2是方程的解
(2)已知(2x+3)/x(x-1)(x+2)=A/x+B/(x-1)+C/(x+2) （A,B,C是常數(shù)）,求A,B,C的值.
(2x+3)/[x(x-1)(x+2)]=A/x+B/(x-1)+C/(x+2)
=[A(x^2+x-2)+B(x^2+2x)+C(x^2-x)]/[x(x-1)(x+2)]
=[(A+B+C)x^2+(A+2B-C)x-2A]/[x(x-1)(x+2)]
可得:
A+B+C=0
A+2B-C=2
-2A=3
A=-3/2
B=5/3
C=-1/6