\(x^5-2022x^4+2020x^3+2020x^2-2020x-2021\)

=\(x^5-x^4-2021x^4+2021x^3-x^3+x^2+2021x^2-2021x+x-1-2020\)

=\(x^4\left(x-1\right)-2021x^3\left(x-1\right)-x^2\left(x+1\right)+2021x\left(x-1\right)+\left(x-1\right)-2020\)

=\(\left(x^4-2021x^3-x^2+2021x+1\right).\left(x-1\right)-2020\)

=\(\left[x^3\left(x-2021\right)-x\left(x-2021\right)+1\right]\left(x-1\right)-2020\)

=\(\left[\left(x^3-x\right).\left(x-2021\right)+1\right]\left(x-1\right)-2020\)*

vì x-2021 luôn bằng 0 \(\Rightarrow\left[\left(x^3-x\right).0+1\right]=1\)

*=1.(2021-1)-2020=0

đây nha bạn //

 dấu hỏi là 57 hoặc 47

Ta có:

 \(1^3=1^2\)

\(1^3+2^3=3^2=\left(1+2\right)^2\)

\(1^3+2^3+3^3=6^2=\left(1+2+3\right)^2\)

\(1^3+2^3+3^3+4^3=10^2=\left(1+2+3+4\right)^2\)

            ....

 Như vậy, dựa theo quy luật trên:

\(\Rightarrow B=\left(1+2+3+...+21\right)^2\)

\(=231^2=53361\)

\(\left(\frac{1}{13}x-7\right)^8+\left(\frac{1}{17}y-7\right)^6=0\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{13}x-7=0\\\frac{1}{17}y-6=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=91\\y=102\end{cases}}\)

\(x+y=91+102=193\)

\(2+2^2+2^3+...+2^{90}+2^{100}\)

\(=\left(2+2^2+2^3+2^4+2^5\right)+2^5\left(2+2^2+2^3+2^4+2^5\right)+...+2^{95}\left(2+2^2+2^3+2^4+2^5\right)\)

\(=62+2^5.62+2^{10}.62+...+2^{95}.62\)

 Mà 62 chia hết cho 31

=> Biểu thức chia hết cho 31

\(2+2^2+2^3+...+2^{90}+2^{100}\)