作业帮两道解分式方程的题~~1. [2(x^2+1)]/(x+1)+[6(x+1)]/(x^2+1)=72. x^4/(x^2+2x+1)-3x^2/(x+1)+2=0过程尽量详细些啊,谢谢!
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/26 01:55:27
![两道解分式方程的题~~1. [2(x^2+1)]/(x+1)+[6(x+1)]/(x^2+1)=72. x^4/(x^2+2x+1)-3x^2/(x+1)+2=0过程尽量详细些啊,谢谢!](/uploads/image/z/7182404-44-4.jpg?t=%E4%B8%A4%E9%81%93%E8%A7%A3%E5%88%86%E5%BC%8F%E6%96%B9%E7%A8%8B%E7%9A%84%E9%A2%98%7E%7E1.++++++%5B2%28x%5E2%2B1%29%5D%2F%28x%2B1%29%2B%5B6%28x%2B1%29%5D%2F%28x%5E2%2B1%29%3D72.++++++x%5E4%2F%28x%5E2%2B2x%2B1%29-3x%5E2%2F%28x%2B1%29%2B2%3D0%E8%BF%87%E7%A8%8B%E5%B0%BD%E9%87%8F%E8%AF%A6%E7%BB%86%E4%BA%9B%E5%95%8A%2C%E8%B0%A2%E8%B0%A2%21)
两道解分式方程的题~~1. [2(x^2+1)]/(x+1)+[6(x+1)]/(x^2+1)=72. x^4/(x^2+2x+1)-3x^2/(x+1)+2=0过程尽量详细些啊,谢谢!
两道解分式方程的题~~
1. [2(x^2+1)]/(x+1)+[6(x+1)]/(x^2+1)=7
2. x^4/(x^2+2x+1)-3x^2/(x+1)+2=0
过程尽量详细些啊,谢谢!
两道解分式方程的题~~1. [2(x^2+1)]/(x+1)+[6(x+1)]/(x^2+1)=72. x^4/(x^2+2x+1)-3x^2/(x+1)+2=0过程尽量详细些啊,谢谢!
令(x^2+1)/(x+1)=a
2a+6/a=7
2a^2-7a+6=0
a=2或a=3/2
(x^2+1)/(x+1)=3/2
3x+3=2x^2+2
2x^2-3x-1=0
x=(3+√17)/4或x=(3-√17)/4
(x^2+1)/(x+1)=2
x^2-2x-1=0
x=1+√2或x=1-√2
经检验,他们都是原方程的根
所以x=(3+√17)/4或x=(3-√17)/4或x=1+√2或x=1-√2
x^4/(x^2+2x+1)=x^4/(x+1)^2=[x^2/(x+1)]^2
令a=x^2/(x+1)
a^2-3a+2=0
a=1,a=2
x^2/(x+1)=1
x^2-x-1=0
x=(1+√5)/2或x=(1-√5)/2
x^2/(x+1)=2
x^2-2x-2=0
x=1+√3或x=1-√3
经检验,他们都是原方程的根
所以x=(1+√5)/2或x=(1-√5)/2或x=1+√3或x=1-√3
呵呵，我会解，可不会把公式写到上面