a. Với điều kiện \(x>0,x\ne1\), ta có:
\(P=\dfrac{x-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{\sqrt{x}\left(x+\sqrt{x}+1\right)}+\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(x-2\sqrt{x}\right)+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+2x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(x+\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\)
b.
\(Q=\dfrac{4x^{2024}+4x^{2023}-2x^{2022}+2023}{2x^{2023}+2x^{2022}-x^{2021}+1}=\dfrac{2x^{2022}\left(2x^2+2x-1\right)+2023}{x^{2021}\left(2x^2+2x-1\right)+1}\)
Do \(x=\dfrac{\sqrt{3}-1}{2}\) là nghiệm của đa thức \(2x^2+2x-1\)
Nên \(Q=\dfrac{2023}{1}=2023\)
