Question

[trần hoàng anh]

(1^+2^2+3^2+....+2021^2)*(91-273:3)

Answer 1

[Sahara]

\(\left(1^2+2^2+3^2+...+2021^2\right)\cdot\left(91-273:3\right)\)
\(=\left(1^2+2^2+3^2+...+2021^2\right)\cdot\left(91-91\right)\)
\(=\left(1^2+2^2+3^2+...+2021^2\right)\cdot0\)
\(=0\)

Answer 2

[Cao Duy Đông]

$\left(1^2+2^2+3^2+...+2021^2\right)\cdot \left(91-273:3\right)$
$=\left(1^2+2^2+3^2+...+2021^2\right)\cdot \left(91-91\right)$
$=\left(1^2+2^2+3^2+...+2021^2\right)\cdot 0$
$=0$