[tex]\frac{2^{2020}+m}{2^{2021}} \to supraunitar\u{a}, \ deci \\ 2^{2020}+m >2^{2021} \\ m>2^{2021}-2^{2020} \\ m>(2-1)\cdot2^{2020} \\ m>2^{2020} \\ m \in (2^{2020}, \ +\infty) \\ Cel \ mai \ mic \ num\u{a}r \ natural \ m \Rightarrow \\m=2^{2020}+1[/tex]
Răspuns: [tex]\red{\bf \underline{m =2^{2020}+1}}[/tex]
Explicație pas cu pas:
Buna !
[tex]\bf \dfrac{2^{2020}+m }{2^{2021} } =fractie~supraunitara~\implies2^{2020}+m>2^{2021}[/tex]
[tex]\bf 2^{2020}+m>2^{2021}[/tex]
[tex]\bf m>2^{2021}-2^{2020}[/tex]
[tex]\bf m>2^{2020}\cdot \Big(2^{2021-2020}-2^{2020-2020}\Big)[/tex]
[tex]\bf m>2^{2020}\cdot \Big(2^{1}-2^{0}\Big)[/tex]
[tex]\bf m>2^{2020}\cdot \big(2-1\big)[/tex]
[tex]\bf m>2^{2020}\cdot 1[/tex]
[tex]\bf m>2^{2020} \implies m \in \Big(2^{2020} , +\infty \Big)[/tex]
[tex]\bf m>2^{2020} \xrightarrow[]{cel~ mai~ mic~ numar}~\red{\boxed{\bf \underline{m =2^{2020}+1}}}[/tex]
Bafta multa !
