\(x=\dfrac{1+\sqrt{3}}{2}\)
\(\Rightarrow2x-1=\sqrt{3}\)
\(\Rightarrow\left(2x-1\right)^2=3\)
\(\Rightarrow4x^2-4x+1=3\)
\(\Rightarrow4x^2-4x-2=0\)
\(\Rightarrow2x^2-2x-1=0\)
Do đó:
\(Q=\left[\left(2x^4-2x^3-x^2\right)-\left(2x^3-2x^2-x\right)-\left(2x^2-2x-1\right)\right]^{2024}\)
\(=\left[x^2\left(2x^2-2x-1\right)-x\left(2x^2-x-1\right)-\left(2x^2-x-1\right)\right]^{2024}\)
\(=\left(x^2.0+x.0+0\right)^{2024}=0\)
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