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\(S=\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}\)
\(\Rightarrow S< \dfrac{3}{10}+\dfrac{3}{10}+\dfrac{3}{10}+\dfrac{3}{10}+\dfrac{3}{10}\)
\(\Rightarrow S< \dfrac{15}{10}< 2\)
Lại có \(S>\dfrac{3}{14}+\dfrac{3}{14}+\dfrac{3}{14}+\dfrac{3}{14}+\dfrac{3}{14}\)
\(\Rightarrow S>\dfrac{15}{14}>1\)
\(\Rightarrow1< S< 2\)
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)
cho s = 3/10+3/11+3/12+3/13+3/14. chứng minh rằng : 1<s<2 . từ đó suy ra s không phải là số tự nhiên
giải\(s>\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}=\frac{15}{15}=1\)
\(s<\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{15}{10}<\frac{20}{10}=2\)
vậy 1<s<2
=>s không thuộc N
\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)
\(\Rightarrow\frac{3}{14}+\frac{3}{14}+\frac{3}{14}+\frac{3}{14}+\frac{3}{14}< S< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}\)
\(\frac{3}{14}\times5< S< \frac{3}{10}\times5\Rightarrow\frac{15}{14}< S< \frac{3}{2}\)
mà \(\frac{15}{14}>1;\frac{3}{2}< 2\Rightarrow1< S< 2\)
\(S=\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}\)
Ta thấy:
\(\dfrac{3}{10}>\dfrac{3}{15}\\\dfrac{3}{11}>\dfrac{3}{15}\\ \dfrac{3}{12}>\dfrac{3}{15}\\ \dfrac{3}{13}>\dfrac{3}{15}\\ \dfrac{3}{14}>\dfrac{3}{15} \)
\(\Rightarrow S=\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}>5\cdot\dfrac{3}{15}\\ S=\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}>1\left(1\right)\)
Mặt khác:
\(\dfrac{3}{10}< \dfrac{3}{9}\\ \dfrac{3}{11}< \dfrac{3}{9}\\ \dfrac{3}{12}< \dfrac{3}{9}\\ \dfrac{3}{13}< \dfrac{3}{9}\\ \dfrac{3}{14}>\dfrac{3}{9}\)
\(\Rightarrow S=\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}< 5\cdot\dfrac{3}{9}\\ S=\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}< \dfrac{5}{3}< 2\left(2\right)\)
Từ (1) và (2) ta có: \(1< S< 2\)