Sửa lại đề nha:

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

Ta có:

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

Lại có:

\(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{15}{10}< \frac{20}{10}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow1< S< 2\left(đpcm\right)\)

Chúc em học tốt!

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

=>3/10+3/11+3/12+3/13+3/14>1=>S>1

S=3/10+3/11+3/12+3/13+3/14 < 3/11+3/11+3/11+3/11+3/11=15/11<2

=>3/10+3/11+3/12+3/13+3/14<2=>S<2

Ta có : 

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

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

\(\Rightarrow S< 1,5\left(1\right)\)

Lại có : 

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

\(\Rightarrow S>\frac{3}{15}.5\)

\(\Rightarrow S>1\left(2\right)\)

Từ ( 1 ) ; ( 2 ) 

\(\Rightarrow1< S< 1,5\)

\(\Rightarrow S\)ko phải là STN 

Hỏa Long Natsu ơi, bạn giải giúp mình một bài nữa đi

Ta có:

\(\frac{1}{m}+\frac{n}{6}=\frac{1}{2}\)

\(\frac{1}{m}=\frac{1}{2}-\frac{n}{6}\)

\(\frac{1}{m}=3-\frac{n}{6}\)

\(\frac{6}{6m}=\frac{\left(3-n\right)m}{6m}\)

\(\left(3-n\right)m=6\Rightarrow\left(-1\right)\left(-6\right)=\left(-2\right).\left(-3\right)=1.6=2.3\)

Đến đây mời bạn xét bảng ><

dễ lắm mai giải cho

nha pn

Tham khảo:

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

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

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

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

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

\(\Rightarrow1< S< 2\)

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

                                                  < 3/10 x 5

                                                 < 3/2 < 2sư

* S = 3/10+3/11+3/12+3/13+3/14 > 3/15+3/15+3/15+3/15+3/15

                                                > 3/15 x 5

                                               > 1

CHỨNG TỎ ........

                                                 > 1

lon de sua 3/1 thanh 3/10

sua 3/1 thanh 3/10

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

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

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: \(1< S< 2\)

sửa đề : \(\frac{9}{10!}+\frac{10}{11!}+\frac{11}{12!}+...+\frac{99}{100!}\)

\(=\frac{10-1}{10!}+\frac{11-1}{11!}+\frac{12-1}{12!}+...+\frac{100-1}{100!}\)

\(=\frac{1}{9!}-\frac{1}{10!}+\frac{1}{10!}-\frac{1}{11!}+\frac{1}{11!}-\frac{1}{12!}+...+\frac{1}{99!}-\frac{1}{100!}\)

\(=\frac{1}{9!}-\frac{1}{100!}< \frac{1}{9!}\left(đpcm\right)\)