Question

已知ɑ²＋4ɑ＋1＝0且ɑ的4次方＋mɑ²＋1／3ɑ³＋mɑ²＋3ɑ＝5,则m＝

Answer 1

a²+4a+1=0

a+1/a= -4

两边平方,

a²+1/a²+2=16

a²+1/a²=14

(a^4+ma²+1)/(3a³+ma²+3a)=5

(a²+1/a²+m) / [ 3(a+1/a)+m] = 5

(14+m)/(-12+m)=5

14+m=-60+5m

m=37/2