Question

数学方程组求解

Answer 1

1/a+(a/4)/b=1，(a/4)/b=1-1/a=(a-1)/a，b=(a/4)/[(a-1)/a]=a²/[4(a-1)]
a²-b²=1，a²-a⁴/[16(a-1)²]=1，a²-a⁴/[16(a²-2a+1)]=1
16a²(a²-2a+1)-a⁴=[16(a²-2a+1)]
16a⁴-32a³+16a²-a⁴=16a²-32a+16
16a⁴-a⁴-32a³+16a²-16a²+32a-16=0
15a⁴-32a³+32a-16=0
a1=-1.00796888548161......
a2=1.50334292355763......
代入b=a²/[4(a-1)]可得到b1和b2的值。