$\begin{array}{l}S=\dfrac3{10}+\dfrac3{11}+\dfrac3{12}+\dfrac3{13}+\dfrac3{14}\\\quad>\dfrac3{15}+\dfrac3{15}+\dfrac3{15}+\dfrac3{15}+\dfrac3{15}\\\quad=\dfrac3{15}\cdot5=1\\\to S>1\quad(1)\\S=\dfrac3{10}+\dfrac3{11}+\dfrac3{12}+\dfrac3{13}+\dfrac3{14}\\\quad<\dfrac39+\dfrac39+\dfrac39+\dfrac39+\dfrac39\\\quad=\dfrac39\cdot5\\\quad=\dfrac{15}9<\dfrac{18}9=2\\\to S<2\quad(2)\\\text{- Từ (1) và (2) $\to 1<S<2$} \end{array}$
A=3/10+3/11+3/12+3/13+3/14 A < 3/10.5
A < 15/10.2
A = 3/10+3/11+3/12+3/13+3/14
A>3/14.5
A<15/14>1
Vậy 1<A<2 (ĐPCM)
$\begin{array}{l}S=\dfrac3{10}+\dfrac3{11}+\dfrac3{12}+\dfrac3{13}+\dfrac3{14}\\\quad>\dfrac3{15}+\dfrac3{15}+\dfrac3{15}+\dfrac3{15}+\dfrac3{15}\\\quad=\dfrac3{15}\cdot5=1\\\to S>1\quad(1)\\S=\dfrac3{10}+\dfrac3{11}+\dfrac3{12}+\dfrac3{13}+\dfrac3{14}\\\quad<\dfrac39+\dfrac39+\dfrac39+\dfrac39+\dfrac39\\\quad=\dfrac39\cdot5\\\quad=\dfrac{15}9<\dfrac{18}9=2\\\to S<2\quad(2)\\\text{- Từ (1) và (2) $\to 1<S<2$} \end{array}$