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Ta có \(\dfrac{6}{15}>\dfrac{6}{16}>...>\dfrac{6}{19}\) nên \(S< \dfrac{6}{15}.5=2\).
Lại có \(S>\dfrac{6}{19}.5>1\) nên \(1< S< 2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) CM : S < 1
Ta có :
6 /15> 6/30
6 /16> 6/30
6/17 > 6/30
6/18 > 6/30
6/19 > 6 / 30
=> S = 6/15 + 6/16 + 6/17 + 6/18 + 6/19 > 6/30 x 5 = 1
=> S > 1 ( 1 )
CM : S < 2
6/16 < 6/15 , 6/17 < 6/15 , 6/18 < 6/15 , 6/19 < 6/15
=> S = 6/15 + 6/16+ 6/17 + 6/18 + 6/19 < 6/15 x 5 = 2
=> S < 2 ( 2 )
Từ ( 1 ) , ( 2 ) => 1 < S < 2
b ) Do 1 < S < 2 => S ko phải STN
Chúc học giỏi !!!
\(S=\frac{6}{15}+\frac{6}{16}+\frac{6}{17}+\frac{6}{18}+\frac{6}{19}>\frac{6}{20}+\frac{6}{20}+\frac{6}{20}+\frac{6}{20}+\frac{6}{20}\)
\(S>\frac{6}{20}\cdot5=\frac{30}{20}\)
\(\Rightarrow S>\frac{3}{2}>1\)
\(S< \frac{6}{14}+\frac{6}{14}+\frac{6}{14}+\frac{6}{14}+\frac{6}{14}\)
\(S< \frac{6}{14}\cdot5=\frac{30}{14}\)
\(S< \frac{15}{7}\Rightarrow S< \frac{14}{7}+\frac{1}{7}\)
\(S< 2+\frac{1}{7}\)
\(\Rightarrow1< \frac{3}{2}< S< 2< 2+\frac{1}{7}\)
\(\Leftrightarrow1< S< 2\Rightarrow S\notin Z\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,1/51 > 1/100
1/52 > 1/100
1/53 > 1/100
...
1/100=1/100
=>H>1/100 + 1/100 + 1/100 +...+1/100
H>50/100=1/2
1/51<1/50
1/52<1/50
....
1/100<1/50
=>H<1/50+1/50+...+1/50
H<50/50=1
Vay1/2<H<1
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(S=\frac{6}{15}+\frac{6}{16}+\frac{6}{17}+\frac{6}{18}+\frac{6}{19}\). Theo như quy tắc đã học ở lớp 5. Ta có:
Các phân số có tử bé hơn mẫu thì phân số đó bé hơn 1
Mà \(\frac{6}{15};\frac{6}{16};\frac{6}{17};\frac{6}{18};\frac{6}{19}\) đều bé hơn 1.
\(\Rightarrow\frac{6}{15}+\frac{6}{16}+\frac{6}{17}+\frac{6}{18}+\frac{6}{19}< 0\RightarrowĐPCM\) (Vì: \(1>\frac{6}{15}>\frac{6}{16}>\frac{6}{17}>\frac{6}{18}>\frac{6}{19}\))
ta có:
6/15+6/16+6/17+6/18+6/19
=31/40+6/17+6/18+6/19
=767/680+6/18+6/19
=1.7777
vậy s không thuộc n
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có:
\(\frac{6}{15}+\frac{6}{16}+...+\frac{6}{19}>\frac{6}{19}.5=\frac{30}{19}>1\)
\(\Rightarrow S>1\)
Ta lại có:
\(\frac{6}{15}+\frac{6}{16}+...+\frac{6}{19}< \frac{6}{15}.5=\frac{30}{15}=2\)
\(\Rightarrow S< 2\)
Vậy, 1 < S < 2
b) \(1< S< 2\Rightarrow S\notin Z\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(S>\frac{6}{20}.5=1,5>1\)
=>S>1
\(S< \frac{6}{15}.5=2\)
=>S<2
Vậy 1<S<2
![](https://rs.olm.vn/images/avt/0.png?1311)
1/5+1/6+1/7+.....+1/17
=5+6+7+....+16+17/5x6x7x8x....x16x17<1
mà 1<2
nên 1/5+1/6+1/7+......+1/16+1/17<2
![](https://rs.olm.vn/images/avt/0.png?1311)
S = 1+ (-3) + 5 + (-7) + .... + 17
S = (-2) x 4 + 17 = 9
S' = -2 + 4 + (-6) + 8 + ..... + (-18)
S' = 2 x 4 + (-18) = -10
Vậy S + S' = -10 + 9 = -1
![](https://rs.olm.vn/images/avt/0.png?1311)
S = (1 - 2 - 3 + 4) + (5 - 6 - 7 + 8) + (9 - 10 - 11 + 12) + (13 - 14 - 15 + 16) + (17 - 18)
= 0 + 0 + 0 + 0 + (-1)
= -1
A = (1 - 2 - 3 + 4) + (5 - 6 - 7 + 8) + ... + (2013 - 2014 - 2015 + 2016) + 2017
= 0 + 0 + ... + 0 + 2017
= 2017
S = (1 - 2 - 3 + 4) + (5 - 6 - 7 + 8) +...... + (13 - 14 - 15 + 16)+ 17 - 18
S = 17 - 18 = -2
A = (1 - 2 - 3 + 4) + (5 - 6 -7 + 8) + ..... + (2013 - 2014 - 2015 + 2016) + 2017
A = 2017
ta có \(S=\frac{6}{15}+\frac{6}{16}+\frac{6}{17}+\frac{6}{18}+\frac{6}{19}\)
\(\Rightarrow S>\frac{6}{20}+\frac{6}{20}+\frac{6}{20}+\frac{6}{20}+\frac{6}{20}\)
\(\Rightarrow S>\frac{30}{20}\)
\(\Rightarrow S>1.5>1\)
\(\Rightarrow s>1\)
Ta có :
\(S=\frac{6}{15}+\frac{6}{16}+\frac{6}{17}+\frac{6}{18}+\frac{6}{19}\)
\(\Rightarrow S< \frac{6}{15}+\frac{6}{15}+\frac{6}{15}+\frac{6}{15}+\frac{6}{15}\)
\(\Rightarrow S< \frac{30}{15}\)
\(\Rightarrow s< 2\)
Vậy \(1< S< 2\)