已知a的平方-3a+1=0,若A=2a的五次方-5a的四次方+2a的三次方-8a的二次方+7a/3a的二次方+3,求A的值

a²-3a=-1
则2a^5-5a^4+2a³-8a²+7a
=2a^5-6a^4+a^4+2a³-8a²+7a
=2a³(a²-3a)+a^4+2a³-8a²+7a
=-2a³+a^4+2a³-8a²+7a
=a^4-3a³+3a³-8a²+7a
=a²×(-1)+3a³-8a²+7a
=3a(a²-3a)+7a
=-3a+7a
=4a
a²+1=3a
3a²+3=9a
所以原式=4a/9a=4/9