\(4^{11}.25^{11}\le2^n.5^n\le20^{12}.5^{12}\)
\(\Rightarrow\left(2^2\right)^{11}.\left(5^2\right)^{11}\le2^n.5^n\le\left(2^2.5\right)^{12}.5^{12}\)
\(\Rightarrow2^{22}.5^{22}\le2^n.5^n\le2^{24}.5^{24}\)
\(\Rightarrow\left(2.5\right)^{22}\le\left(2.5\right)^n\le\left(2.5\right)^{24}\)
\(\Rightarrow22\le n\le24\Rightarrow n\in\left\{22;23;24\right\}\left(n\in N\right)\)
Ta có 411.2511=(4.25)11=10011=1022
2n.5n=(2.5)n=10n
2012.512=(20.5)12=10012=1024
⇒ 1022≤10n<1024
⇒ 22≤n<24
⇒ nϵ {22;23}
=>10011≤ 10n≤10012
=>(102)11≤10n≤(102)12
=>1022≤10n≤1024
=>nϵ(22;23;24)
hình như Lê Minh Đức thiếu 24 ở phần đáp án thì phải
4^11.25^11<hoặc=2^n.5^n<hoặc=20^12.5^12 tìm n
=>(4x25)\(^{11}\)\(\le\)(2x5)\(^n\)\(\le\)(20x5)\(^{12}\)
=>100\(^{11}\)\(\le\)10\(^n\)\(\le\)100\(^{12}\)
=>10\(^{110}\)\(\le\)10\(^n\)\(\le\)10\(^{120}\)
=>n\(\in\)\(\left\{110;111;...;120\right\}\)