题目已知aaa是方程x2−2020x+4=0x^{2}-2020x+4=0x2−2020x+4=0的一个解，则a2−2019a+8080a2+4+6a^{2}-2019a+\dfrac{8080}{a^{2}+4}+6a2−2019a+a2+48080+6的值为()(\quad)()A.202220222022B.202120212021C.202020202020D.201920192019知识点：一元二次方程的解章节：未标注
