Question

已知a²–1=3a+1 则a³–a–7a+1=

Answer 1

问：已知a²–1=3a+1 则a³–a–7a+1=

答：您好亲亲已知a²–1=3a+1 则a³–a–7a+1= 。a²-a=1得出a²-a-1=0a³+a²-3a+7 = a³-a²-a+2a²-2a+7 = a(a²-a-1) + 2(a²-a) + 7 =2+7 = 9

答：相关资料：我来补答a²-3a+1=0,得：a²+1=3a3a﹙³﹢¹﹚-7a³-3a²+a／a²+1=3a^4-7a³-3a²+a/(a²+1）=3a²（a²-3a+1）+2a³-6a²+a/(3a)=0+2a(a²-3a+1）-2a+1/3=0+0-2a+1/3=1/3-2a a²-3a+1=0利用求根公式算出a=（3±√5）/2,带入上面的式子就可以了.