\(x^2+xy+y^2=x^2y^2\)

\(\Leftrightarrow4x^2+4xy+4y^2=4x^2y^2\)

\(\Leftrightarrow4x^2+8xy+4y^2=4x^2y^2-4xy+1-1\)

\(\Leftrightarrow\left(2x+2y\right)^2=\left(2xy-1\right)^2-1\)

\(\Leftrightarrow\left(2xy-1+2x+2y\right)\left(2xy-1-2x-2y\right)=1\)

\(A=1+4+4^2+...+4^{2021}\)

\(=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)...+\left(4^{2019}+4^{2020}+4^{2021}\right)\)

\(=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...+4^{2019}\left(1+4+4^2\right)\)

\(=21\left(1+4^3+...+4^{2019}\right)⋮21\)

\(=21+21.4^3+...+21.4^{2019}\)

Dùng định lý talet chứng minh là được.

Ta có CD//EF

\(\Rightarrow\widehat{C}+\widehat{F}=180^0\)

Ta lại có: \(\Rightarrow\widehat{C}-\widehat{F}=66^0\)nên ta có hệ

\(\left\{{}\begin{matrix}\widehat{C}+\widehat{F}=180^0\\\widehat{C}-\widehat{F}=66^0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\widehat{C}=123^0\\\widehat{F}=57^0\end{matrix}\right.\)

Thế n = 2 ta được:

\(\dfrac{2.3.5}{6}=5\)

Xong

Theo đề bài ta có:

\(x_0+x_1=-a_1;x_0.x_1=b_1\)

\(x_0+x_2=-a_2;x_0.x_2=b_2\)

............................................

\(x_0+x_{2022}=-a_{2022};x_0.x_{2022}=b_{2022}\)

Ta có:

\(x_0+\alpha=x_0+\dfrac{x_1+x_2+...+x_{2022}}{2022}=\dfrac{\left(x_0+x_1\right)+\left(x_0+x_2\right)+...+\left(x_0+x_{2022}\right)}{2022}=-\dfrac{a_1+a_2+...+a_{2022}}{2022}\)\(x_0\alpha=x_0\dfrac{x_1+x_2+...+x_{2022}}{2022}=\dfrac{x_0x_1+x_0x_2+...+x_0x_{2022}}{2022}=\dfrac{b_1+b_2+...+b_{2022}}{2022}\)

Từ đây ta có được \(x_0;\alpha\)là 2 nghiệm của phương trình 

\(x^2+\dfrac{a_1+a_2+...+a_{2022}}{2022}x+\dfrac{b_1+b_2+...+b_{2022}}{2022}=0\)

Hỏi câu nào thế?

\(\sqrt{4}< \sqrt{6+\sqrt{6+\sqrt{...+\sqrt{6}}}}< \sqrt{6+\sqrt{6+\sqrt{...+\sqrt{9}}}}\)

\(\Leftrightarrow2< \sqrt{6+\sqrt{6+\sqrt{...+\sqrt{6}}}}< 3\)

Vậy phần nguyên là 2

Lấy 2 phương trình nhân vế theo vế được:

\(x^3y^3=\left(xy+4\right)\left(3xy-4\right)\)

Đặt \(xy=t\) thì được:

\(t^3=\left(t+4\right)\left(3t-4\right)\)

\(\left(t-4\right)\left(t^2+t-4\right)=0\)

Làm nốt

Câu này dùng bất: \(\sqrt{ab}\le\dfrac{a+b}{2}\)

Áp dụng bài toán được:

\(A=\dfrac{1}{\sqrt{1.1999}}+\dfrac{1}{\sqrt{2.1998}}+...\dfrac{1}{\sqrt{1999.1}}\)

\(>\dfrac{1}{\dfrac{1+1999}{2}}+\dfrac{1}{\dfrac{2+1998}{2}}+...+\dfrac{1}{\dfrac{1999+1}{2}}\)

\(=\dfrac{1}{1000}+\dfrac{1}{1000}+...+\dfrac{1}{1000}\)

Làm nốt