1/(x+1)+1/(x+5)=1/(x+2)+1/(x+4)解方程

来源：百度知道 编辑：UC知道 时间：2024/05/08 10:35:42
详细过程

1/(x+1)+1/(x+5)=1/(x+2)+1/(x+4)
1/(x+1)-1/(x+2)=1/(x+4)-1/(x+5)
[(x+2)-(x+1)]/[(x+1)(x+2)]=[(x+5)-(x+4)]/[(x+4)(x+5)]
1/[(x+1)(x+2)]=1/[(x+4)(x+5)]
x^2+3x+2=x^2+9x+20
6x=-18
x=-3

1/(x+1)+1/(x+5)=1/(x+2)+1/(x+4)

1/(x+1)-1/(x+2)=1/(x+4)-1/(x+5)

1/(x+1)(x+2)=1/(x+4)(x+5)

(x+1)(x+2)=(x+4)(x+5)

x^2+3x+2=x^2+9x+20

6x=-18

x=-3

等式两边移项通分得(x+2)(x+3)=(x+4)(x+5) 接下来解一元二次方程即可

1/(x+1)+1/(x+5)=1/(x+2)+1/(x+4)
1/(x+1)(x+5) = 1/(x+2)(x+4)
1/(x^2 + 6x + 5) = 1/(x^2 + 6x + 8)
x^2 + 6x + 5 = x^2 + 6x + 8