Question

已知a的平方+3a+1=0，求a的平方/a的4次方+a的平方+1的值。拜托了各位 谢谢

Answer 1

a^2 + 3a + 1 = 0 ==> a^2 = -3a - 1 a^2 + 3a = -1 a^2 + 1 = -3a a^2/a^4 + a^2 + 1 = 1/a^2 + a^2 + 1 = 1/a^2 - 3a = 1/(-3a-1) - 3a = -1/(3a + 1) - 3a = - [1 + 3a(3a+1)]/(3a+1) = - [1 + 9a^2 + 3a] / (3a+1) = - [1 + 9(-3a-1) + 3a] / (3a+1) = - [1 - 27a - 9 + 3a] / (3a+1) = - [-24a - 8] / (3a + 1) = (24a+8) / (3a+1) = 8(3a+1) / (3a+1) = 8