Answer:

`5(3x+2)=4x+1`

\(\Leftrightarrow15x+10=4x+1\)

\(\Leftrightarrow15x-4x=1-10\)

\(\Leftrightarrow11x=-9\)

\(\Leftrightarrow x=-\frac{9}{11}\)

\(\left(x-3\right)\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}}\)

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

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

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

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

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

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

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:

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

\(=231^2=53361\)

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

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

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

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