Ta có: \(x=\sqrt{\dfrac{1}{2\sqrt{3}-2}-\dfrac{3}{2\sqrt{3}+2}}\)
\(\Rightarrow x^2=\dfrac{1}{2\sqrt{3}-2}-\dfrac{3}{2\sqrt{3}+2}\)
\(=\dfrac{1}{2\left(\sqrt{3}-1\right)}-\dfrac{3}{2\left(\sqrt{3}+1\right)}\)
\(=\dfrac{\sqrt{3}+1}{2\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\dfrac{3\left(\sqrt{3}-1\right)}{2\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=\dfrac{\sqrt{3}+1-3\sqrt{3}+3}{2.\left(3-1\right)}\)
\(=\dfrac{-2\sqrt{3}+4}{4}\)
\(=\dfrac{\left(\sqrt{3}-1\right)^2}{4}\)
\(\Rightarrow x=\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{4}}=\dfrac{\left|\sqrt{3}-1\right|}{2}=\dfrac{\sqrt{3}-1}{2}\) (do \(x\ge0\))
\(\Leftrightarrow2x=\sqrt{3}-1\)
\(\Leftrightarrow2x+1=\sqrt{3}\)
\(\Leftrightarrow\left(2x+1\right)^2=3\)
\(\Leftrightarrow4x^2+4x+1=3\)
\(\Leftrightarrow4x^2+4x-2=0\)
\(\Leftrightarrow2x^2+2x-1=0\)
Lại có: \(P=\dfrac{4\left(x+1\right)^{2024}-2x^{2023}+2x+3}{2x^2+3x}\)
\(=\dfrac{2x^{2023}\left[2x\left(x+1\right)-1\right]+2x+3}{2x^2+3x}\)
\(=\dfrac{2x^{2023}\left(2x^2+2x-1\right)+2x+3}{2x^2+3x}\)
\(=\dfrac{2x+3}{2x^2+3x}\left(1\right)\) (do \(2x^2+2x-1=0\))
Thay \(x=\dfrac{\sqrt{3}-1}{2}\) vào \(\left(1\right)\) ta được:
\(P=\dfrac{2.\dfrac{\sqrt{3}-1}{2}+3}{2.\left(\dfrac{\sqrt{3}-1}{2}\right)^2+3.\dfrac{\sqrt{3}-1}{2}}=1+\sqrt{3}\)
