+TH₁: \(M>1\)

 Ta có: \(\frac{3}{10}>\frac{3}{15}=\frac{1}{5}\)

            \(\frac{3}{11}>\frac{3}{15}=\frac{1}{5}\)

            \(\frac{3}{12}>\frac{3}{15}=\frac{1}{5}\)

            \(\frac{3}{13}>\frac{3}{15}=\frac{1}{5}\)

            \(\frac{3}{14}>\frac{3}{15}=\frac{1}{5}\)

 \(\Rightarrow M=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}>\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=5.\frac{1}{5}=1\)

 \(\Rightarrow M>1\) 1

+TH₂: \(M< 2\)

 Ta có: \(\frac{3}{10}< \frac{3}{9}=\frac{1}{3}\)

           \(\frac{3}{11}< \frac{3}{9}=\frac{1}{3}\)

           \(\frac{3}{12}< \frac{3}{9}=\frac{1}{3}\)

           \(\frac{3}{13}< \frac{3}{9}=\frac{1}{3}\)

           \(\frac{3}{14}< \frac{3}{9}=\frac{1}{3}\)

 \(\Rightarrow M=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< \frac{1}{3}\frac{1}{3}\frac{1}{3}\frac{1}{3}\frac{1}{3}=\frac{5}{3}< \frac{6}{3}=2\)

 \(\Rightarrow M< 2\) 2

Từ 1 và 2 \(\Rightarrow1< M< 2\)hay M không phải là số tự nhiên

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

\(\Rightarrow S< \dfrac{3}{10}+\dfrac{3}{10}+\dfrac{3}{10}+\dfrac{3}{10}+\dfrac{3}{10}\)

\(\Rightarrow S< \dfrac{15}{10}< 2\)

Lại có \(S>\dfrac{3}{14}+\dfrac{3}{14}+\dfrac{3}{14}+\dfrac{3}{14}+\dfrac{3}{14}\)

\(\Rightarrow S>\dfrac{15}{14}>1\)

\(\Rightarrow1< S< 2\)

Tham khảo:

Ta có: \(\dfrac{3}{10}>\dfrac{3}{15}\)

\(\dfrac{3}{11}>\dfrac{3}{15}\)

\(\dfrac{3}{12}>\dfrac{3}{15}\)

\(\dfrac{3}{13}>\dfrac{3}{15}\)

\(\dfrac{3}{14}>\dfrac{3}{15}\)

Do đó: \(\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}>\dfrac{3}{15}+\dfrac{3}{15}+\dfrac{3}{15}+\dfrac{3}{15}+\dfrac{3}{15}=1\)

hay 1<S(1)

Ta có: \(\dfrac{3}{11}< \dfrac{3}{10}\)

\(\dfrac{3}{12}< \dfrac{3}{10}\)

\(\dfrac{3}{13}< \dfrac{3}{10}\)

\(\dfrac{3}{14}< \dfrac{3}{10}\)

Do đó: \(\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}< \dfrac{3}{10}+\dfrac{3}{10}+\dfrac{3}{10}+\dfrac{3}{10}=\dfrac{12}{10}\)

\(\Leftrightarrow S< \dfrac{15}{10}=\dfrac{3}{2}< 2\)(2)

Từ (1) và (2) suy ra 1<S<2(đpcm)

Bạn làm hay quá

S> 3/15 .5

S>1

S< 3/10x5=3/2 <2 

cậu giải chi tiết hơn đc ko

S=3/10+3/11+3/12+3/13+3/14
S<3/10.5

S<15/10<2

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

S>3/14.5

S<15/14>1

Vậy 1<s<2 (ĐPCM)
_Yên tâm cách giải=))Con này đội tuyển toán 6
Nhớ k