求解一个一元五次方,102=8/(1+y)+8/(1+y)^2+8/(1+y)^3+8/(1+y)^4+108/(1+y)^5y=?

Question

求解一个一元五次方,102=8/(1+y)+8/(1+y)^2+8/(1+y)^3+8/(1+y)^4+108/(1+y)^5 y=?

Answer 1

设t=1/(y+1),则原式化为：
108t^5+8t^4+8t^3+8t^2+8t-102=0
此方程只有一个实根：
t=0.930184127443755
所以有：y=1/t-1=0.075055971