Đáp án:
$B=2020$
Giải thích các bước giải:
ĐK: ${a_1},{a_2},{a_3},...,{a_{2019}},{a_{2020}} \ne 0$
Ta có:
$\begin{array}{l}
\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{a_2}}}{{{a_3}}} = ... = \dfrac{{{a_{2019}}}}{{{a_{2020}}}} = \dfrac{{{a_{2020}}}}{{{a_1}}}\\
\Rightarrow \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{a_2}}}{{{a_3}}} = ... = \dfrac{{{a_{2019}}}}{{{a_{2020}}}} = \dfrac{{{a_{2020}}}}{{{a_1}}} = \dfrac{{{a_1} + {a_2} + ... + {a_{2019}} + {a_{2020}}}}{{{a_2} + {a_3} + ... + {a_{2020}} + {a_1}}} = 1\\
\Rightarrow {a_1} = {a_2} = {a_3} = ... = {a_{2019}} = {a_{2020}}
\end{array}$
Khi đó:
$\begin{array}{l}
B = \dfrac{{{{\left( {{a_1} + {a_2} + ... + {a_{2020}}} \right)}^2}}}{{a_1^2 + a_2^2 + ... + a_{2020}^2}}\\
= \dfrac{{{{\left( {2020{a_1}} \right)}^2}}}{{2020a_1^2}}\\
= \dfrac{{{{2020}^2}a_1^2}}{{2020a_1^2}}\\
= 2020
\end{array}$
Vậy $B=2020$
