3/10>3/15

3/11>3/15

3/12>3/15

3/13>3/15

3/14>3/15

=>S>3/15*5=15/15=1

3/11<3/10

3/12<3/10

3/13<3/10

3/14<3/10

=>3/11+3/12+3/13+3/14+3/10<3/10*5=15/10=3/2<2

=>1<S<2

\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}>\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}=\frac{15}{15}=1\)

\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< \frac{3}{9}+\frac{3}{9}+\frac{3}{9}+\frac{3}{9}+\frac{3}{9}=\frac{15}{9}< \frac{18}{9}=2\)

Suy ra đpcm.

S < 3/10 +3/10 +3/10  + 3/10 + 3/10 = 15/ 10  < 20/10 =2

bạn my hang làm chuẩn

my hang làm chuẩn ko cần chỉnh luôn

bấm vào chữ Đúng 0 sẽ hiện ra kết quả 

Ta có: S =3/10+3/11+3/12+3/13+3/14 = 3.(1/10+1/11+1/12+1/13+1/14) > 3.(1/15 + 1/15 + 1/15 + 1/15 + 1/15) = 3.5/15 = 1 => S > 1       (1)

         S=3/10+3/11+3/12+3/13+3/14 = 3.(1/10+1/11+1/12+1/13+1/14) < 3.(1/10 + 1/10 + 1/10 + 1/10 + 1/10) = 3.5/10 = 3/2<2 =>S <2  (2)

 Từ (1) va (2)

=> 1 < S < 2 (đpcm).

Chúc bạn học tập tốt :)

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

=> S < 5 x 3/10

=> S < 1,5

=> S <  2

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

=> S > 5 x 3/14

=> S > 1,07.......

Mà 1 < 1,07 < S < 1,5 < 2

=> 1 < S < 2

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

\(S>\frac{3}{14}.5\)

\(S>\frac{15}{14}>1\)(1)

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

\(S< 5.\frac{3}{10}\)

\(S< \frac{15}{10}< 2\)(2)

Từ (1) và (2) => 1 < S < 2 => S không là số nguyên tố (đpcm)

a/ta có 

3/10>3/14

3/11>3/14

...

=>S>5.3/14

=>s>15/14>1

b,ta có 

3/10>3/11

3/10>3/12

....

3/10>3/14

=>s<5.3/10<15/10<2

S>3/15+3/15+3/15+3/15+3/15=15/15=1

S<3/10+3/10+3/10+3/10+3/10=15/10=3/2<2

⇒S ko ϵ N

ta có : S > 3/14 + 3/14 + 3/14 + 3/14 + 3/14

S > 15/14 > 14/14 = 1

S < 3/10 + 3/10 + 3/10 + 3/10 + 3/10

S < 15/10 < 20/10 = 2

vậy 1 < S < 2