ta có (2+\(\sqrt{3}\))^9=140452 ===> (2+\(\sqrt{3}\))^2021 là số nguyên  

(2-\(\sqrt{3}\))^2021 là số thập phân gần tiến tới 0                                                                                            vậy (2+\(\sqrt{3}\))^2021 +(2-\(\sqrt{3}\))^2021 k thể là số nguyên

Đề bài đúng. Dùng công thức (a+b)^n và (a-b)^n
 

:v e xin kiếu

Em tham khảo nhé

https://hoc24.vn/cau-hoi/cho-xsqrtx22021ysqrty220212021tinh-axy.332667728355

\(f\left(x\right)=ax^2+bx+2020\\ \Leftrightarrow f\left(\sqrt{3}-1\right)=a\left(4-2\sqrt{3}\right)+b\left(\sqrt{3}-1\right)+2020=2021\\ \Leftrightarrow4a-2a\sqrt{3}+b\sqrt{3}-b-1=0\\ \Leftrightarrow\left(4a-b-1\right)-\sqrt{3}\left(2a-b\right)=0\\ \Leftrightarrow4a-b-1=\sqrt{3}\left(2a-b\right)\)

Vì a,b hữu tỉ nên \(4a-b-1;2a-b\) hữu tỉ

Mà \(\sqrt{3}\) vô tỉ nên \(\sqrt{3}\left(2a-b\right)\) hữu tỉ khi \(2a-b=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}4a-b-1=0\\2a-b=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=1\end{matrix}\right.\)

\(\Leftrightarrow f\left(1+\sqrt{3}\right)=\dfrac{1}{2}\left(4+2\sqrt{3}\right)+1+\sqrt{3}+2020=2023+2\sqrt{3}\)

\(x=\dfrac{\sqrt[3]{\left(2+\sqrt{3}\right)^3}\left(2-\sqrt{3}\right)}{\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}}=\dfrac{1}{\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}}\)

Đặt \(A=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)\(\Leftrightarrow A^3=18+3\sqrt[3]{\left(9-4\sqrt{5}\right)\left(9+4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\\ \Leftrightarrow A^3=18+3A\sqrt[3]{1}\\ \Leftrightarrow A^3-3A-18=0\\ \Leftrightarrow A=3\\ \Leftrightarrow X=\dfrac{1}{3}\\ \Leftrightarrow Q=\left[3\left(\dfrac{1}{3}\right)^3-\left(\dfrac{1}{3}\right)^2-1\right]^{2021}=\left(\dfrac{1}{9}-\dfrac{1}{9}-1\right)^{2021}=\left(-1\right)^{2021}=-1\)

`x=\root{3}{4(\sqrt5+1)}-\root{3}{4(\sqrt5-1)}`

`<=>x^3=4(sqrt5+1)-4(\sqrt5-1)-3\root{3}{16(5-1)}(\root{3}{4(\sqrt5+1)}-\root{3}{4(\sqrt5-1)})`

`<=>x^3=4\sqrt5+4-4sqrt5+4-3\root{3}{64}x`

`<=>x^3=8-12x`

`<=>x^3+12x-8=0`

`=>P=(x^3+12-8-1)^2021=(-1)^2021=-1`

*Có gì khum hiểu comment bên dưới.

\(\dfrac{2}{3}-\left|\dfrac{3}{4}\right|+\sqrt{\dfrac{25}{9}}-\left(\dfrac{2021}{2022}\right)^0=\dfrac{2}{3}-\dfrac{3}{4}+\dfrac{5}{3}-1=\dfrac{7}{12}\)

\(=\dfrac{2}{3}-\dfrac{3}{4}+\dfrac{5}{3}-1=\dfrac{7}{12}\)