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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 !!!
\(S=\frac{6}{15}+\frac{6}{16}+\frac{6}{17}+\frac{6}{18}+\frac{6}{19}>\frac{6}{20}+\frac{6}{20}+\frac{6}{20}+\frac{6}{20}+\frac{6}{20}\)
\(S>\frac{6}{20}\cdot5=\frac{30}{20}\)
\(\Rightarrow S>\frac{3}{2}>1\)
\(S< \frac{6}{14}+\frac{6}{14}+\frac{6}{14}+\frac{6}{14}+\frac{6}{14}\)
\(S< \frac{6}{14}\cdot5=\frac{30}{14}\)
\(S< \frac{15}{7}\Rightarrow S< \frac{14}{7}+\frac{1}{7}\)
\(S< 2+\frac{1}{7}\)
\(\Rightarrow1< \frac{3}{2}< S< 2< 2+\frac{1}{7}\)
\(\Leftrightarrow1< S< 2\Rightarrow S\notin Z\)
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
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 \(1< S< 2\)
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) \(1< S< 2\Rightarrow S\notin Z\)
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
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
Ta có \(S>\frac{6}{20}.5=1,5>1\)
=>S>1
\(S< \frac{6}{15}.5=2\)
=>S<2
Vậy 1<S<2
Đặt A = \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+....+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}\)
\(A=\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}\right)+\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{14}\right)+\left(\frac{1}{15}+\frac{1}{16}+...+\frac{1}{19}\right)\)
\(\Rightarrow A< \left(\frac{1}{5}+...+\frac{1}{5}\right)+\left(\frac{1}{10}+...+\frac{1}{10}\right)+\left(\frac{1}{15}+...+\frac{1}{15}\right)\)
\(\Rightarrow A< \frac{1}{5}\cdot5+\frac{1}{10}\cdot5+\frac{1}{15}\cdot5\)
\(\Rightarrow A< 1+\frac{1}{2}+\frac{1}{3}\)
\(\Rightarrow A< \frac{11}{6}< 2\)
\(\Rightarrow A< 2\left(đpcm\right)\)
\(S=\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+...+\frac{6}{29.32}\)
\(S=2.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\right)\)
\(S=2.\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(S=2.\frac{15}{31}\Rightarrow S=\frac{15}{16}< 1\)
\(S=\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+...+\frac{6}{29.32}\)
\(S=\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+...+\left(\frac{1}{29}-\frac{1}{32}\right)\)
\(S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\)
\(S=\frac{1}{2}-\frac{1}{32}\)
\(S=\frac{17}{32}< 1\)
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 \(1< S< 2\)