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) 

\(S>\frac{3}{15}+\frac{3}{15}+...\frac{3}{15}\left(5\right)số\frac{3}{15}\)

\(=\frac{15}{15}=1\)

\(S>\frac{3}{10}+...+\frac{3}{10}\left(5so\right)\)

\(=\frac{15}{10}< \frac{20}{10}=2\)

\(=>1< P< 2\)

Vậy P không phải là số tự nhiên.

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

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

> \(3.\left(\frac{1}{14}+\frac{1}{14}+\frac{1}{14}+\frac{1}{14}+\frac{1}{14}\right)\)

= \(3\left(\frac{1}{14}.5\right)\)

= \(3.\frac{5}{14}\)

= \(\frac{15}{14}\)> 1 

=> S > \(\frac{15}{14}\)>1

=> S > 1 (1)

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

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

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

= \(3.\left(\frac{1}{10}.5\right)\)

= \(3.\frac{1}{2}\)

= \(\frac{3}{2}\)<2

=> S < \(\frac{3}{2}\)< 2

=> S < 2 (2)

Từ (1) và (2) ta có 

1 < S < 2

=> S không là số tự nhiên 

3/10>3/15               3/11>3/15

3/12>3/15                 3/13>3/15

            3/14>3/15                                       

vẬY 3/10 + 3/11 + 3/12+ 3/13+3/14 > 3/15+3/15+3/15+3/15+3/15=15/15=1

           => 3/10+3/11+3/12+3/13+3/14>1       (1)

3/10<3/9                3/11<3/9           3/12<3/9            3/13<3/9            3/14<3/9

VẬY 3/10+3/11+3/12+3/13+3/14<3/9+3/9+3/9+3/9+3/9=15/9    MÀ 15/9<18/9=2

3/10+3/11+3/12+3/13+3/14<2          (2)

   TỪ 1 VÀ 2 =>   1<S<2

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

Bài làm

a) \(-\frac{3}{7}+\frac{3}{4}:\frac{3}{14}\)

= \(-\frac{3}{7}+\frac{3}{4}.\frac{14}{3}\)

= \(-\frac{3}{7}+\frac{7}{2}\)

\(=-\frac{7}{14}+\frac{49}{14}\)

\(=\frac{42}{14}=3\)

b) \(5-\frac{7}{39}:\frac{7}{13}+\frac{8}{9}:4\)

\(=5=\frac{7}{39}.\frac{13}{7}+\frac{8}{9}.\frac{1}{4}\)

\(=5-\frac{1}{3}+\frac{2}{9}\)

\(=\frac{45}{9}-\frac{3}{9}+\frac{2}{9}\)

\(=\frac{44}{9}\)

c) \(\left(\frac{5}{12}:\frac{11}{6}+\frac{5}{12}:\frac{11}{5}\right)-\frac{-7}{12}\)

\(=\left(\frac{5}{12}.\frac{6}{11}+\frac{5}{12}.\frac{5}{11}\right)+\frac{7}{12}\)

\(=\left[\frac{5}{12}\left(\frac{6}{11}+\frac{5}{11}\right)\right]+\frac{7}{12}\)

\(=\frac{5}{12}+\frac{7}{12}\)

\(=\frac{12}{12}=1\)

d) \(-\frac{5}{9}+\frac{14}{9}\left(\frac{3}{4}-\frac{2}{5}\right):49\)

\(=-\frac{5}{9}+\frac{14}{9}\left(\frac{15}{20}-\frac{8}{20}\right):49\)

\(=-\frac{5}{9}+\frac{14}{9}.\frac{7}{20}.\frac{1}{49}\)

\(=-\frac{5}{9}+\frac{7}{9}.\frac{7}{10}.\frac{1}{7.7}\)

\(=-\frac{5}{9}+\frac{1}{90}\)

\(=-\frac{50}{90}+\frac{1}{90}=-\frac{49}{90}\)