1.(x+7)/(x+6) + (x+9)/(x+8) = (x+10)/(x+9) + (x+6)/(x+5)2.已知：-x+13/（x-5)(x+3) =A/(x-5)-B/(x+3).求A,B3.已知abc=1,求a/(ab+a+1) + b/(bc+b+1) + c/(ca+c+1)

1.(x+7)/(x+6) + (x+9)/(x+8) = (x+10)/(x+9) + (x+6)/(x+5)2.已知：-x+13/（x-5)(x+3) =A/(x-5)-B/(x+3).求A,B3.已知abc=1,求a/(ab+a+1) + b/(bc+b+1) + c/(ca+c+1)
1.(x+7)/(x+6) + (x+9)/(x+8) = (x+10)/(x+9) + (x+6)/(x+5)
2.已知：-x+13/（x-5)(x+3) =A/(x-5)-B/(x+3).求A,B
3.已知abc=1,求a/(ab+a+1) + b/(bc+b+1) + c/(ca+c+1)

1.(x+7)/(x+6) + (x+9)/(x+8) = (x+10)/(x+9) + (x+6)/(x+5)2.已知：-x+13/（x-5)(x+3) =A/(x-5)-B/(x+3).求A,B3.已知abc=1,求a/(ab+a+1) + b/(bc+b+1) + c/(ca+c+1)
1 原方程可化为：1+1/(x+6)+1+1/(x+8)=1+1/(x+9)+1+1/(x+5)
既1/(x+6)-1/(x+5)=1/(x+9)-1/(x+8)
通分得-1/(x+6)(x+5)=-1/(x+9)(x+8)
既(x+6)(x+5)=(x+9)(x+8)
解得x=-7
2 方程右边同分得（-x+13）/（x-5)(x+3)=[(A-B)X+3A+5B]/(x-5)(x+3)
既-x+13=(A-B)x+3A+5B
A-B=-1 3A+5B=13,A=1 B=2
3 a=1/bc b=1/ac
a/(ab+a+1)=a/(1/c+a+1)=ac/(1+ac+c)
b/(bc+b+1)=(1/ac)/(1/a+1/ac+1)=1/(ac+c+1)
故a/(ab+a+1) + b/(bc+b+1) + c/(ca+c+1)
=ac/(1+ac+c)+1/(ac+c+1)+c/(ca+c+1)
=(c+1+ac)/(ac+c+1)=1
看看我的第一题的做法,比他们都简单.

1.(x+7)/(x+6) + (x+9)/(x+8) = (x+10)/(x+9) + (x+6)/(x+5)
思路：经观察，原分式方程的四个分式中，分子都比分母大1，所以可以先分拆化简，比较方便。
因为(x+7/x+6)=1+1/(x+6)，其它同理，所以可得：
1+1/(x+6)+1+1/(x+8)=1+1/(x+9)+1+1/(x+5)
1/(x+6...

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1.(x+7)/(x+6) + (x+9)/(x+8) = (x+10)/(x+9) + (x+6)/(x+5)
思路：经观察，原分式方程的四个分式中，分子都比分母大1，所以可以先分拆化简，比较方便。
因为(x+7/x+6)=1+1/(x+6)，其它同理，所以可得：
1+1/(x+6)+1+1/(x+8)=1+1/(x+9)+1+1/(x+5)
1/(x+6)+1/(x+8)=1/(x+9)+1/(x+5)
1/(x+6)+1/(x+8)-[1/(x+9)+1/(x+5)]=0
通分，得：
[(x+8)+(x+6)]/[(x+6)(x+8)]-[(x+5)+(x+9)]/[(x+9)(x+5)]=0
(2x+14)/(x²+14x+48)-(2x+14)/(x²+14x+45)=0
(2x+14)[1/(x²+14x+48)-1/(x²+14x+45)]=0
可见，中括号内的两个分式，分子相同，分母不等，所以差不等于0，只能是：
2x+14=0
x=－7
代入原方程检验，x=－7符合要求。
2.已知：-x+13/（x-5)(x+3) =A/(x-5)-B/(x+3).求A，B
二边同乘上(x-5)(x+3):
-x+13=A(x+3)-B(x-5)=(A-B)x+(3A+5B)
左右二边一一对应得:
A-B=-1
3A+5B=13
解得:A=1,B=2.
3.已知abc=1，求a/(ab+a+1) + b/(bc+b+1) + c/(ca+c+1)
abc=1
所以b=1/ac
ab=1/c
bc=1/a
所以
原式=a/(1/c+a+1)+(1/ac)/(1/a+1/ac+1)+c/(ac+c+1)
第一个式子上下同乘c
第二个式子上下同乘ac
所以=ac/(ac+c+1)+1/(ac+c+1)+c/(ac+c+1)
=(ac+c+1)/(ac+c+1)
=1

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1，展开 在合并同类项
2 两边通分 比较系数
3 [a/(ab+a+1)+b/(bc+b+1)+c/(ac+c+1)
=ac/(1+ac+c)+ab/(1+ab+a)+c/(ac+c+1)
=(ac+c)/(ac+c+1)+1/(ac+c+1)
=(ac+c+1)/(ac+c+1)=1

我说,这位同学,这是你们的家庭作业是不是啊?呵呵,,,,,

第一题目：将X+5设成M
(m+2)/(m+1)-(m+1)/m=(m+5)/(m+4)-(m+4)/(m+3)
M=-2
X+5=-2
X=-7

(1)1+1/(x+6)+1+1/(x+8)=1+1/(x+9)+1+1/(x+5)
1/(x+6)+1/(x+8)=1/(x+9)+1/(x+5)
设x+7=n
1/(n-1)+1/(n+1)=1/(n+2)+1/(n-2)
2n/(n^2-1)=2n/(n^2-4)
n=0=x+7
x=-7
(2)x移动到等号右边，右边的分...

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(1)1+1/(x+6)+1+1/(x+8)=1+1/(x+9)+1+1/(x+5)
1/(x+6)+1/(x+8)=1/(x+9)+1/(x+5)
设x+7=n
1/(n-1)+1/(n+1)=1/(n+2)+1/(n-2)
2n/(n^2-1)=2n/(n^2-4)
n=0=x+7
x=-7
(2)x移动到等号右边，右边的分式移动到等号左边，得到
((B-A)x+13-5B-3A)/(x-5)(x+3) = x
所以B-A=1 且 13-5B-3A=0
A=1,B=2
可能数算的有问题，不过方法没错
(3)
=ac/(1+ac+c)+ab/(1+ab+a)+c/(ac+c+1)
=(ac+c)/(ac+c+1)+1/(ac+c+1)
=(ac+c+1)/(ac+c+1)=1

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1.(x+7)/(x+6) + (x+9)/(x+8) = (x+10)/(x+9) + (x+6)/(x+5)
1+1/(x+6)+1+1/(x+8)=1+1/(x+9)+1+1/(x+5)
(2x+14)/(x+6)(x+8)=(2x+14)/(x+9)(x+5)
(x+6(x+8)不可能=(x+9)(x+5)
分母不能相等，只有 2x+14=0

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1.(x+7)/(x+6) + (x+9)/(x+8) = (x+10)/(x+9) + (x+6)/(x+5)
1+1/(x+6)+1+1/(x+8)=1+1/(x+9)+1+1/(x+5)
(2x+14)/(x+6)(x+8)=(2x+14)/(x+9)(x+5)
(x+6(x+8)不可能=(x+9)(x+5)
分母不能相等，只有 2x+14=0
x=-7 经检验x=-7是原方程的根
2.已知：-x+13/（x-5)(x+3) =A/(x-5)-B/(x+3).求A，B
3.已知abc=1，求a/(ab+a+1) + b/(bc+b+1) + c/(ca+c+1)

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1、变型为：1+1/(x+6) + 1+1/(x+8) = 1+1/(x+9) + 1+1/(x+5)
1/(x+6) + 1/(x+8) = 1/(x+9) + 1/(x+5)
(2x+14)/(x+6) (x+8) = (2x+14)/(x+9) + 1/(x+5)
当X不等于-7时，得到(x+5)(x+9)=(x+6)(x+8) 不成立，所以x=-7

1.(x+7)/(x+6)-(x+6)/(x+5) = (x+10)/(x+9)- (x+9)/(x+8)
[(x+7)(x+5)-(x+6)(x+6)]/(x+6)(x+5) = [(x+10)(x+8)-(x+9)(x+9)]/(x+9)(x+8)
-1/(x+6)(x+5) = -1/(x+9)(x+8)
(x+6)(x+5) =(x+9)(x+8)
11x...

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1.(x+7)/(x+6)-(x+6)/(x+5) = (x+10)/(x+9)- (x+9)/(x+8)
[(x+7)(x+5)-(x+6)(x+6)]/(x+6)(x+5) = [(x+10)(x+8)-(x+9)(x+9)]/(x+9)(x+8)
-1/(x+6)(x+5) = -1/(x+9)(x+8)
(x+6)(x+5) =(x+9)(x+8)
11x+30=17x+72
6x = -42
x= -7
2.已知：-x+13/（x-5)(x+3) =A/(x-5)-B/(x+3).求A，B
-x+13/（x-5)(x+3) =[A(x+3)-B(x-5)]/(x-5)(x+3)
-x+13=A(x+3)-B(x-5)
AX-BX+X+3A+5B-28=0
有无穷解
3.已知abc=1，求a/(ab+a+1) + b/(bc+b+1) + c/(ca+c+1)
a/(ab+a+1) + b/(bc+b+1) + c/(ca+c+1)=1/(b+bc+1) + 1/(c+ac+1) + 1/(a+ab+1)

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