Ta có : \(x=\frac{\sqrt{5}-1}{2}\Rightarrow2x=\sqrt{5}-1\)
\(\Leftrightarrow2x+1=\sqrt{5}\)
\(\Rightarrow\left(2x+1\right)^2=5\)
\(\Rightarrow4x^2+4x+1=5\Rightarrow x^2+x-1=0\)
Khi đó ta có :
\(B=\left(4x^5+4x^4-5x^3+2x-2\right)^2+2021\)
\(=\left[\left(4x^5+4x^4-4x^3\right)-\left(x^3+x^2-x\right)+\left(x^2+x-1\right)-1\right]^2+2021\)
\(=\left[4x^3\left(x^2-x+1\right)-x\left(x^2+x-1\right)+\left(x^2+x-1\right)-1\right]^2+2021\)
\(=\left(-1\right)^2+2021=2022\)
Vậy \(B=2022\)