Question

数学初三一元二次方程求解

Answer 1

解：∵a是方程x²+1=3x的根，
∴a²+1=3a
∴a²-3a=-1

a³-2a²-5a+1
=a³-3a²+a²-5a+1
=a(a²-3a)+a²-5a+1
=a×(-1)+a²-5a+1
=a²-6a+1
=a²-3a-3a+1
=-1-3a+1
=-3a

a²+1=3a
∴原分式＝(-3a)/(3a)=-1

Answer 2

a的3次-2a的2次-5a+1可改写成 a的3次-3a的平方-2a
分母 为3a
最后得 -1

Answer 3

解：x^2+1=3x 所以：a^2+1=3a 即：a^2-3a=-1
a^3-2a^2-5a+1/a^2+1 = a^3-3a^2+a^2-3a-2a+1/a^2-3a+3a+1 = a*(a^2-3a) + (a^2-3a) -2a+1/(a^2-3a)+3a+1 = a*(-1) + (-1) -2a+1/(-1) +3a+1 = -3a/3a = -1 ，答案为 -1 。此题注意转化，不知楼主是否满意！