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\(TH1;n=3k\)\(\Rightarrow10^n+18n-1=\)\(10^{3k}+18.3k-1=1000^k+54k-1\equiv1+54k-1\left(mod27\right)\equiv0\left(mod27\right)\left(1\right)\)
\(TH2;n=3k+1\Rightarrow10^n+18n-1=10^{3k+1}+18.\left(3k+1\right)-1\)\(=10^{3k}.10+18.\left(3k+1\right)-1=1000^k.10+54k+18-1\)\(\equiv1.10+54k+17\left(mod27\right)\equiv54k+27\left(mod27\right)\equiv0\left(mod27\right)\left(2\right)\)
\(TH3;n=3k+2\Rightarrow10^n+18n-1=10^{3k+2}+54k+36-1\)\(=1000^{3k}.100+54k+35\equiv1.100+54k+35\left(mod27\right)\)\(\equiv54k+135\left(mod27\right)\equiv0\left(mod27\right)\left(3\right)\)\(Từ\left(1\right);\left(2\right);\left(3\right)\Rightarrow10^n+18n-1⋮27,\forall n\in N\left(ĐPCM\right)\)
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Ta có: \(\frac{12}{13}=1-\frac{1}{13}\) ; \(\frac{22}{23}=1-\frac{1}{23}\)
Do \(\frac{1}{13}>\frac{1}{23}\)nên \(1-\frac{1}{13}< 1-\frac{1}{23}\)
Vậy \(\frac{12}{13}< \frac{22}{23}\)
\(\frac{12}{13}=1-\frac{1}{13};\frac{22}{23}=1-\frac{1}{23}\)
Có \(1-\frac{1}{13}< 1-\frac{1}{23}\Rightarrow\frac{12}{13}< \frac{22}{23}\)
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(a,b).[a,b]=a.b
=>(a,b)=135:45
=>(a,b)=3
ta có ƯCLN(a,b)=3
a=3.a' b=3.b'
ta có
a.b=135
=>3.a'.3.b'=135
=>9.a'.b'=135
=>a'.b'=15
a' | 1 | 3 | 5 | 15 |
b' | 15 | 5 | 3 | 1 |
=>
a | 3 | 9 | 15 | 45 |
b | 45 | 15 | 9 | 3 |
k cho mk nha