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1)
Ta có :
2300 = ( 23 )100 = 8100
3200 = ( 32 )100 = 9100
vì 8100 < 9100 nên 2300 < 3200
2)
Ta có :
523 = 522 . 5
vì 522 . 5 < 522 . 6 nên 523 < 6 . 522
Ta có : \(S=1+2+2^2+...+2^{2019}\)
\(\Leftrightarrow2S=2+2^2+2^3+....+2^{2020}\)
\(\Leftrightarrow S=2^{2020}-1\)
Ta thấy : \(5.2^{2018}=\left(4+1\right).2^{2018}=2^{2020}+2^{2018}>2^{2020}-1\)
Do đó : \(S< 5\cdot2^{2018}\)
Ta có S = 1 + 2 + 22 + ... + 22019
=> 2S = 2 + 22 + 23 + ... + 22020
Lấy 2S trừ S theo vế ta có :
2S - S = (2 + 22 + 23 + ... + 22020) - (1 + 2 + 22 + ... + 22019)
S = 22020 - 1
Lại có : 5 . 2018 = (22 + 1).22018 = 22020 + 22018
Vì 22020 - 1 < 22020 + 22018
=> S < 5.22018
Vậy S < 5.22018
\(P=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+\frac{4}{5^5}+...+\frac{11}{5^{12}}\)
\(\Rightarrow\)\(5P=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+\frac{4}{5^4}+...+\frac{11}{5^{11}}\)
\(\Rightarrow\)\(4P=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+...+\frac{1}{5^{11}}-\frac{1}{5^{12}}\)
\(\Rightarrow\)\(20P=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{10}}-\frac{1}{5^{11}}\)
\(\Rightarrow\)\(16P=1-\frac{1}{5^{11}}+\frac{1}{5^{12}}-\frac{1}{5^{11}}\)\(< 1\)
\(\Rightarrow\)\(P< \frac{1}{16}\)
P/s: nguyên tác: https://olm.vn/thanhvien/nhatphuonghocgiot
1. \(n^2+n-17\)là bội của n+5\(\Leftrightarrow\)\(n^2+n-17\)chia hết n+5
Ta có \(n^2+n-17⋮n+5\)
\(\Rightarrow n^2+n+5-22⋮n+5\)
\(\Rightarrow22⋮n+5\)
\(\Rightarrow n+5\inƯ\left(22\right)\)
\(\Rightarrow n+5\in\left(1;2;11;22;-1;-2;-11;-22\right)\)
\(\Rightarrow n\in\left(-4;-3;6;17;-6;-7;-16;-27\right)\)
1,
ta có : \(\frac{\overline{abab}}{\overline{cdcd}}=\frac{\overline{abab}:101}{\overline{cdcd}:101}=\frac{\overline{ab}}{\overline{cd}}\) ; \(\frac{\overline{ababab}}{\overline{cdcdcd}}=\frac{\overline{ababab}:10101}{\overline{cdcdcd}:10101}=\frac{\overline{ab}}{\overline{cd}}\)
Vậy \(\frac{\overline{abab}}{\overline{cdcd}}=\frac{\overline{ababab}}{\overline{cdcdcd}}\)
2,
\(\frac{1}{2}.\frac{1}{b}=\frac{2}{4}\)
\(\Rightarrow\frac{1.1}{2.b}=\frac{2}{4}\)
\(\Rightarrow\frac{1}{2.b}=\frac{1}{2}\)
\(\Rightarrow2.b=2\)
\(\Rightarrow b=2:2=1\)
\(\frac{abab}{cdcd}=\frac{abab:101}{cdcd:101}=\frac{ab}{cd}\)
mà \(\frac{ababab}{cdcdcd}=\frac{ababab:10101}{cdcdcd:10101}=\frac{ab}{cd}\)
=> \(\frac{abab}{cdcd}=\frac{ababab}{cdcdcd}\)
vậy...
câu 2
\(\frac{1}{2}.\frac{1}{b}=\frac{2}{4}\\ \Rightarrow\frac{1}{b}=\frac{2}{4}:\frac{1}{2}=1\\ \Rightarrow b=1\)
vậy....