\(S=2^0+2^1+2^2+...+2^7\)

\(\Rightarrow S=\left(2^0+2^1\right)+2^2\left(2^0+2^1\right)+...+2^6\left(2^0+2^1\right)\)

\(\Rightarrow S=3+2^2.3+...+2^6.3\)

\(\Rightarrow S=3\left(1+2^2+...+2^6\right)⋮3\)

\(\Rightarrow dpcm\)

\(S=\frac{5}{20}+\frac{5}{21}+\frac{5}{22}+...+\frac{5}{49}\)

\(S>5\left(\frac{1}{49}+\frac{1}{49}+...+\frac{1}{49}\right)\)(30 số hạng \(\frac{1}{49}\))

\(\Leftrightarrow S>5.\frac{30}{49}\)

\(\Leftrightarrow S>\frac{150}{49}=3\frac{3}{49}\)

\(\Rightarrow S>3\)

\(\Rightarrow S>\frac{3}{49}\)

Vậy \(3< S\)  (1)

Ta lại có: \(S< 5.\left(\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\right)\)(30 số hạng)

\(S< \frac{30}{20}.5=\frac{150}{20}=\frac{15}{2}=7\frac{1}{2}\)

\(\Rightarrow S< 7< 8\)

\(\Rightarrow S< \frac{1}{2}\)

Vậy \(S< 8\) (2)

Từ (1) và (2) ta có đpcm

thank you very much

khi minh vua dang

\(S=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{95}+2^{96}\right)\\ S=\left(1+2\right)\left(2+2^3+...+2^{95}\right)\\ S=3\left(2+2^3+...+2^{95}\right)⋮3\left(1\right)\\ S=\left(2+2^2\right)+2^3\left(1+2^2+...+2^{93}\right)\\ S=8+8\left(1+2^2+...+2^{93}\right)⋮8\left(2\right)\\ \left(1\right)\left(2\right)\Rightarrow S⋮24\)

do \(\frac{5}{20}< 1;\frac{5}{21}< 1;\frac{5}{22}< 1;\frac{5}{23}< 1;\frac{5}{24}< 1\)

\(\Rightarrow\frac{5}{20}+\frac{5}{21}+\frac{5}{22}+\frac{5}{23}+\frac{5}{24}< 1\)

Vậy S < 1

Mk nghĩ thế bn ạ

Ai thấy tớ đúng ủng hộ nha

Ta có: \(\frac{5}{20}>\frac{5}{25}\)

\(\frac{5}{21}>\frac{5}{25}\)

\(\frac{5}{22}>\frac{5}{25}\)

\(\frac{5}{23}>\frac{5}{25}\)

\(\frac{5}{24}>\frac{5}{25}\)

=> \(S>\frac{5}{25}+\frac{5}{25}+\frac{5}{25}+\frac{5}{25}+\frac{5}{25}=5\cdot\frac{5}{25}=\frac{25}{25}=1\)

Vậy S > 1

Ta có :

\(\frac{5}{20}>\frac{5}{25}\)

\(\frac{5}{21}>\frac{5}{25}\)

\(\frac{5}{22}>\frac{5}{25}\)

\(\frac{5}{23}>\frac{5}{25}\)

\(\frac{5}{24}>\frac{5}{25}\)

\(\Rightarrow S>\frac{5}{25}+\frac{5}{25}+\frac{5}{25}+\frac{5}{25}+\frac{5}{25}=5\cdot\frac{5}{25}=\frac{25}{25}=1\)

Vậy \(S>1\)

Ta có :

\(\frac{5}{20}>\frac{5}{25}\)

\(\frac{5}{21}>\frac{5}{25}\)

\(\frac{5}{22}>\frac{5}{25}\)

\(\frac{5}{23}>\frac{5}{25}\)

\(\frac{5}{24}>\frac{5}{25}\)

\(\Rightarrow\frac{5}{20}+\frac{5}{21}+\frac{5}{22}+\frac{5}{23}+\frac{5}{24}>5.\frac{5}{25}=1\)

\(\Rightarrow\frac{5}{20}+\frac{5}{21}+\frac{5}{22}+\frac{5}{23}+\frac{5}{24}>1\)

ta có S=5/20+5/21+5/22+5/23+5/24>5/25+5/25+5/25+5/25+5/25=5/25*5=1

=>đpcm

\(S=1+2+2^2+2^3+2^4+...+2^{2011}\)

\(\Rightarrow S=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+...+2^{2009}\left(1+2+2^2\right)\)

\(\Rightarrow S=7+2^3.7+...+2^{2009}.7\)

\(\Rightarrow S=7\left(1+2^3+...+2^{2009}\right)⋮7\)

\(\Rightarrow dpcm\)

ta thấy : 1/21>1/33;...1/30>1/33

Vậy 1/21+..+1/30>1/33+...+1/33(10 lần 1/33)

1/3=11/33

mà 1/33+..+1/33(10 lần 1/33) =10/33

Suy ra S>1/33+..+1/33(10 lần 1/33)>1/3

Vậy S>1/3

nhớ k nha bạn 

viết lôn nha câu đầu la .. 1/30.>1/33

Thanks

\(S=\left(1+2\right)+...+2^6\left(1+2\right)=3\left(1+...+2^6\right)⋮3\)