Question

解分式方程:1/(x+2) + 1/(x+7) = 1/(x+3) + 1/(x+6)

Answer 1

通分
(x+7+x+2)/(x+2)(x+7)=(x+6+x+3)/(x+3)(x+6)
(2x+9)/(x^2-9x+14)-(2x+9)/(x^2+9x+18)=0
(2x+9)[1/(x^2-9x+14)-1/(x^2+9x+18)]=0
因为x^2-9x+14不等于x^2+9x+18
所以1/(x^2-9x+14)-1/(x^2+9x+18)不等于0
所以2x+9=0
x=-9/2
分式方程要检验
经检验
x=-9/2是方程的解

Answer 2

1/(x+2) + 1/(x+7) = 1/(x+3) + 1/(x+6)
1/(x+2) - 1/(x+3) = 1/(x+6)- 1/(x+7)
通分 :
1/[(x+2)(x+3)]=1/[(x+6)(x+7)]
(x+2)(x+3)=(x+6)(x+7)
x^2+5x+6=x^2+13x+42
8x=-36

x=-9/2

Answer 3

(x+7+x+2)/(x+2)(x+7)=(x+6+x+3)/(x+3)(x+6)
(2x+9)/(x^2-9x+14)-(2x+9)/(x^2+9x+18)=0
(2x+9)[1/(x^2-9x+14)-1/(x^2+9x+18)]=0
因为x^2-9x+14不等于x^2+9x+18
所以1/(x^2-9x+14)-1/(x^2+9x+18)不等于0
所以2x+9=0
x=-9/2