Question

Vă rog frumos ajutați-ma , am nevoie de ea până mâine dim. . Mulțumesc! ​

Answer 1

$\it b=\sqrt{2020+1+2+2^2+2^3+\ ...\ +2^{2019}-2019}=\\ \\ =\sqrt{2020-2019+(1+2+2^2+2^3+\ ...\ + 2^{2019})}=\sqrt{1+\dfrac{2^{2020}-1}{2-1}}=\\ \\ \\ =\sqrt{1+2^{2020}-1}=\sqrt{2^{2020}}=\sqrt{(2^{1010})^2}=2^{1010$