(1)你的题应是（2x-3)/(x^2-x)=A/(x-1)+B/x

　　如果是这样,可得,(2x-3)/(x^2-x)=(Ax+Bx-B)/(X^2-X)

　　这个方程相当于恒等式,也就是说各项系数分别相等.

　　所以,A+B=2

　　(2)题就是(3a^2-5ab-2b^2)/(2a^2+3ab-5b^2)

　　由(a-b)/b=1/2,得,2a=3b

　　(3a^2-5ab-2b^2)/(2a^2+3ab-5b^2)

　　＝[（3a+b)(a-2b)]/[(2a+5b)(a-b)]

　　将2a=3b代入上式得：[（9b/2+b)(3b/2-2b)]/[(3b+5b)(3b/2-b)]=[(11b/2)*(-b/2)]/(8b*b/4)=-11/16.

　　(3)题应是x^2/(x-1)-(1+1/(x^2-x))=x^2/(x-1)-(x^2-x+1)/(x^2-x)=(x^3-x^2+x-1)/(x^2-x)=[(x^2+1)(x-1)]/[x*(x-1)]=(x^2+1)/x

　　由,x^2-4x+1=0,得,x^2+1=4x即(x^2+1)/x=4

　　所以x^2/(x-1)-(1+1/(x^2-x))=4

　　（4）题应是a/(a+1)+b/(b+1)+c/(c+1)=

　　[1-1/(a+1)]+[1-1/(b+1)]+[1-(c+1)]=3-[1/(a+1)+1/(b+1)+1/(c+1)]

　　又1/(a+1）＝1/[x/(y+z)+1]=1/[(x+y+z)/(y+z)]=

　　(y+z)/(x+y+z)

　　1/(b+1)=(z+x)/(x+y+z)

　　1/(c+1)=(x+y)/(x+y+z)

　　所以

　　a/(a+1)+b/(b+1)+c/(c+1)=3-[1/(a+1)+1/(b+1)+1/(c+1)]＝3-〔(y+z)/(x+y+z)+(z+x)/(x+y+z)+(x+y)/(x+y+z)〕＝3-（y+z+z+x+x+y)/(x+y+z)=1