Question

解方程(x+1)/(x+2)+(x-6)/(x+7)=(x+2)/(x+3)+(x+5)/(x+6) 求高手解答，下午要啊

搞错了，是x+6

Answer 1

怀疑如果中间(x-6)是(x+6)
(x+1)/(x+2)+(x+6)/(x+7)=(x+2)/(x+3)+(x+5)/(x+6)
(x+1)/(x+2)-(x+2)/(x+3)=(x+5)/(x+6)-(x+6)/(x+7)
[(x+1)(x+3)-(x+2)²]/[(x+2)(x+3)]=[(x+5)(x+7)-(x+6)²]/[(x+6)(x+7)]
[x²+4x+3-x²-4x-4]/(x²+5x+6) = [x²+12x+35-x²-12X-36] / [x²+13x+42]
-1/(x²+5x+6) = -1/(x²+13x+42)
则 x²+5x+6=x²+13x+42
8x=-36
x=-9/2

Answer 2

解：原式化简为：1-1/(x+2)+1-1/(x+7)=1-1/(x+3)+1-1/(x+6)
即：1/(x+2)-1/(x+3)=1/(x+6)-1/(x+7)
即：(x+2)(x+3)=(x+6)(x+7)
得：x= -9/2

Answer 3

左边第二个分子是x+6还是x-6

Answer 4

[x(x+3)][(x+1)(x+2)]=360 (x^2+3x)(x^2+3x+2)=360 设t=x^2+3x+1,则有(t-1)(t+1)=360,解得t^2=361,t1=19,t2=-19 则有x^2+3x+1=19,解得x1=3,x2=-6 或x^2+3x+1=-19,此时(x+3/2)^2+71/4=0,冲突,舍去 以是x1=3,x2=-6 建议考虑，仅供参考
cawxytxozn74324307532012/5/10 14:36:50

Answer 5

X=1/3