方式方程1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)=1/(x+4)的解是x=

来源:学生作业帮助网 编辑:六六作业网 时间:2024/05/03 17:57:22
方式方程1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)=1/(x+4)的解是x=方式方程1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)=

方式方程1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)=1/(x+4)的解是x=
方式方程1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)=1/(x+4)的解是x=

方式方程1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)=1/(x+4)的解是x=
1/(x+2)(x+1)+1/(x+3)(x+2)+1/(x+3)(x+4)=1/(x+4)
1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)=1/(x+4)
1/(x+1)=2/(x+4)
x=2

x=2
通分 化简之后 3(x+2)(x+3)=(x+1)(x+2)(x+3)
约分 最后得出 结果 x=2