作业帮将f(x)=x/(x^2-2x-3)展开①f(x)=x/(x^2-2x-3)=1/4[3/(x-3)+1/(x+1)]=(-1/4)*[1/(1-x/3)]+(1/4)*[1/(1+x)]=(-1/4)∑(n=0)(x/3)^n+(1/4)∑(n=0)(-x)^n②f(x)=x/(x^2-2x-3)=(-x/12)*[1/(1-x/3)]-(x/4)[1/(1+x)]=(-1/4)∑(n=0)(x/3)^(n+1)+(1/4)∑(n=0)(-x)^(n+1)
来源：学生作业帮助网 编辑：作业帮 时间：2024/07/27 13:54:50
![将f(x)=x/(x^2-2x-3)展开①f(x)=x/(x^2-2x-3)=1/4[3/(x-3)+1/(x+1)]=(-1/4)*[1/(1-x/3)]+(1/4)*[1/(1+x)]=(-1/4)∑(n=0)(x/3)^n+(1/4)∑(n=0)(-x)^n②f(x)=x/(x^2-2x-3)=(-x/12)*[1/(1-x/3)]-(x/4)[1/(1+x)]=(-1/4)∑(n=0)(x/3)^(n+1)+(1/4)∑(n=0)(-x)^(n+1)](/uploads/image/z/1137155-59-5.jpg?t=%E5%B0%86f%28x%29%3Dx%2F%28x%5E2-2x-3%29%E5%B1%95%E5%BC%80%E2%91%A0f%28x%29%3Dx%2F%28x%5E2-2x-3%29%3D1%2F4%5B3%2F%28x-3%29%2B1%2F%28x%2B1%29%5D%3D%28-1%2F4%29%2A%5B1%2F%281-x%2F3%29%5D%2B%281%2F4%29%2A%5B1%2F%281%2Bx%29%5D%3D%28-1%2F4%29%E2%88%91%28n%3D0%29%28x%2F3%29%5En%2B%281%2F4%29%E2%88%91%28n%3D0%29%28-x%29%5En%E2%91%A1f%28x%29%3Dx%2F%28x%5E2-2x-3%29%3D%28-x%2F12%29%2A%5B1%2F%281-x%2F3%29%5D-%28x%2F4%29%5B1%2F%281%2Bx%29%5D%3D%28-1%2F4%29%E2%88%91%28n%3D0%29%28x%2F3%29%5E%28n%2B1%29%2B%281%2F4%29%E2%88%91%28n%3D0%29%28-x%29%5E%28n%2B1%29)
将f(x)=x/(x^2-2x-3)展开①f(x)=x/(x^2-2x-3)=1/4[3/(x-3)+1/(x+1)]=(-1/4)*[1/(1-x/3)]+(1/4)*[1/(1+x)]=(-1/4)∑(n=0)(x/3)^n+(1/4)∑(n=0)(-x)^n②f(x)=x/(x^2-2x-3)=(-x/12)*[1/(1-x/3)]-(x/4)[1/(1+x)]=(-1/4)∑(n=0)(x/3)^(n+1)+(1/4)∑(n=0)(-x)^(n+1)
将f(x)=x/(x^2-2x-3)展开
①f(x)=x/(x^2-2x-3)=1/4[3/(x-3)+1/(x+1)]=(-1/4)*[1/(1-x/3)]+(1/4)*[1/(1+x)]=(-1/4)∑(n=0)(x/3)^n+(1/4)∑(n=0)(-x)^n
②f(x)=x/(x^2-2x-3)=(-x/12)*[1/(1-x/3)]-(x/4)[1/(1+x)]=(-1/4)∑(n=0)(x/3)^(n+1)+(1/4)∑(n=0)(-x)^(n+1)
为什么有2个答案啊 哪里出错了
将f(x)=x/(x^2-2x-3)展开①f(x)=x/(x^2-2x-3)=1/4[3/(x-3)+1/(x+1)]=(-1/4)*[1/(1-x/3)]+(1/4)*[1/(1+x)]=(-1/4)∑(n=0)(x/3)^n+(1/4)∑(n=0)(-x)^n②f(x)=x/(x^2-2x-3)=(-x/12)*[1/(1-x/3)]-(x/4)[1/(1+x)]=(-1/4)∑(n=0)(x/3)^(n+1)+(1/4)∑(n=0)(-x)^(n+1)
①的首项为0,去掉首项即为
(-1/4)∑(n=1)(x/3)^n+(1/4)∑(n=1)(-x)^n.
②就是
(-1/4)∑(n=1)(x/3)^n+(1/4)∑(n=1)(-x)^n.
两者是一样的.
，第二个n初值为-1，或第一个n初值为1，这样就是一样了，只是形式不同罢了。
x/[(x-3)(x+1)]