Ta có : \(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)

                \(=3.\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}\right)\)

                 \(>3.\left(\frac{1}{15}+\frac{1}{15}+\frac{1}{15}+\frac{1}{15}+\frac{1}{15}\right)\)

                  \(=3.\frac{1}{3}=1\)

=> S > 1 (1)

Ta có : 

: \(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)

                \(=3.\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}\right)\)

                \(< 3.\left(\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}\right)\)

                \(=3.\frac{1}{2}=\frac{3}{2}< \frac{4}{2}=2\)    

=> S < 2 (2)

Từ (1) và (2) => 1 < S < 2 (đpcm) 

\(E=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+...+\frac{1}{73\cdot75}\)

\(E=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)

\(\Rightarrow E=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)=\frac{1}{2}\cdot\frac{2}{75}=\frac{1}{75}\)

\(F=\frac{15}{90\cdot94}+\frac{15}{94\cdot98}+...+\frac{15}{146\cdot150}\)

\(F=\frac{15}{4}\cdot\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+...+\frac{1}{146}-\frac{1}{150}\right)\)

\(\Rightarrow F=\frac{15}{4}\cdot\left(\frac{1}{90}-\frac{1}{150}\right)=\frac{15}{4}\cdot\frac{1}{225}=\frac{1}{60}\)

\(G=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)

\(G=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)

\(G=\frac{5}{4\cdot7}+\frac{5}{7\cdot10}+\frac{5}{10\cdot13}+...+\frac{5}{25\cdot28}\)

\(G=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)

\(\Rightarrow G=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{3}\cdot\frac{3}{14}=\frac{5}{14}\)

sao nhiều vậy bạn 

Ta có : 

\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)

\(2S=6+3+\frac{3}{2}+...+\frac{3}{2^8}\)

\(2S-S=\left(6+3+\frac{3}{2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\)

\(S=6-\frac{3}{2^9}\)

\(S=\frac{2^{10}.3-3}{2^9}\)

Vậy \(S=\frac{2^{10}.3-3}{2^9}\)

vận dụng 3S lên

xong tìm S nha bn ok

tại k có thời gian nên chỉ giúp thế thôi

Bài 1 mik học xong quên hết òi (mấy bài kia là hok biết luôn :V)

Ta có 

S=3/10 + 3/11 + 3/12 + 3/13 + 3/14

Suy ra S<[3/10+3/10+3/10+3/10+3/10]

Suy ra S<2/3

MÀ 2/3 < 4/5 suy ra S<4/5

Ta lại có :

S=3/10 + 3/11 +3/12 +3/13 +3/14

Suy ra S>[3/14 + 3/14 + 3/14 + 3/14 + 3/14]

Suy ra S> 15/14

MÀ 15/14 > 3/5 suy ra S>3/5

Từ hai thứ ta chứng minh thì ta có: 3/5<S<4/5

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+....+\frac{1}{97.100}=\frac{0,33.x}{2009}\)

\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{97}-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)

\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)

\(\Leftrightarrow\frac{1}{3}\cdot\frac{99}{100}=\frac{0,33.x}{2009}\)

\(\Leftrightarrow\frac{33}{100}=\frac{0,33.x}{2009}\)

\(\Leftrightarrow x=\frac{0,33\times100}{0,33}=100\)

1.

\(3^{500}=\left(3^5\right)^{100}\)

\(7^{300}=\left(7^3\right)^{100}\)

\(3^5< 7^3\Leftrightarrow3^{500}< 7^{300}\)

\(3^{500}=\left(3^5\right)^{100}\)

\(7^{300}=\left(7^3\right)^{100}\)

3⁵ < 7³ => 3⁵⁰⁰ <7³⁰⁰

M = 10/56 + 10/140 + 10/260 +...+ 10/1400
M = 5/28 + 5/70 + 5/130 + ... + 5/700
M = 5/4.7 + 5/7.10 + 5/10.13 + ... + 5/25.28
M = 5.1/3.(3/4.7 + 3/7.10 + 3/10.13 + ... + 3/25.28)
M = 5/3.(1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13 + ... + 1/25 - 1/28)
M = 5/3.(1/4 - 1/28)
M = 5/3.3/14
M = 5/14