Không dùng máy tính bỏ túi hãy chứng minh S >1
S = \(\dfrac{5}{20}\)+\(\dfrac{5}{21}\)+\(\dfrac{5}{22}\)+\(\dfrac{5}{23}\)+\(\dfrac{5}{24}\)
K
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LK
17 tháng 5 2017
Ta có:5/20>5/25
5/21>5/25
5/22>5/25
5/23>5/25
5/24>5/25
=>S=5/20+5/21+5/22+5/23+5/24>5/25+5/25+5/25+5/25+5/25=1
=>5/20+5/21+5/22+5/23+5/24>1
19 tháng 5 2017
DỄ
DO: 5/20 <1
5/21<1
5/22<1
5/23<1
5/24<1
=> 5/20+5/21+5/22+5/23+5/24<1
hay S<1 ( ĐPCM)
ĐÚNG NÈ ỦNG HỘ
H9
HT.Phong (9A5)
CTVHS
1 tháng 3 2023
\(F=\dfrac{5}{6}+6\dfrac{5}{6}\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)
\(F=\dfrac{5}{6}+\dfrac{41}{6}\left(\dfrac{225}{20}-\dfrac{37}{4}\right):\dfrac{25}{3}\)
\(F=\dfrac{5}{6}+\dfrac{41}{6}.2.\dfrac{3}{25}\)
\(F=\dfrac{5}{6}+\dfrac{41}{25}.\dfrac{3}{25}\)
\(F=\dfrac{5}{6}+\dfrac{41}{25}\)
\(F=\dfrac{371}{150}\)
H9
HT.Phong (9A5)
CTVHS
1 tháng 3 2023
\(D=\left(\dfrac{136}{15}-\dfrac{28}{5}+\dfrac{62}{10}\right)\times\dfrac{21}{24}\)
\(D=\left(\dfrac{272}{30}-\dfrac{168}{30}+\dfrac{186}{30}\right)\times\dfrac{21}{24}\)
\(D=\dfrac{290}{30}\times\dfrac{21}{24}\)
\(D=\dfrac{29}{3}\times\dfrac{7}{8}\)
\(D=\dfrac{203}{24}\)
O
1
15 tháng 6 2023
S=(1/31+1/32+...+1/40)+(1/41+...+1/50)+(1/51+...+1/60)
=>S>1/40*10+1/50*10+1/60*10=3/5
S=(1/31+1/32+...+1/40)+(1/41+...+1/50)+(1/51+...+1/60)
=>S<1/30*10+1/40*10+1/50*10=4/5
=>3/5<S<4/5
28 tháng 4 2023
1/31>1/40
1/32>1/40
...
1/40=1/40
=>1/31+1/32+...+1/40>1/40*10=1/4
1/41>1/50
1/42>1/50
...
1/50=1/50
=>1/41+1/42+...+1/50>10/50=1/5
1/51>1/60
1/52>1/60
...
1/60=1/60
=>1/51+1/52+...+1/60>10/60=1/6
=>S>1/4+1/5+1/6=3/5
1/31<1/30
1/32<1/30
...
1/40<1/30
=>1/31+1/32+...+1/40<1/30*10=1/3
1/41<1/40
1/42<1/40
...
1/50<1/40
=>1/41+1/42+...+1/50<10/40=1/4
1/51<1/50
1/52<1/50
...
1/60<1/50
=>1/51+1/52+...+1/60<10/50=1/5
=>S<1/3+1/4+1/5=4/5
Ta có:\(\dfrac{1}{20}>\dfrac{1}{21}>\dfrac{1}{22}>\dfrac{1}{23}>\dfrac{1}{24}>\dfrac{1}{25}\)
=>S=\(\dfrac{5}{20}+\dfrac{5}{21}+\dfrac{5}{22}+\dfrac{5}{23}+\dfrac{5}{24}=5\left(\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+\dfrac{1}{24}\right)>5\cdot\left(\dfrac{1}{25}+\dfrac{1}{25}+\dfrac{1}{25}+\dfrac{1}{25}+\dfrac{1}{25}\right)\)
=>S>\(5\cdot\dfrac{5}{25}\)
=>S>1(đpcm)
Cảm ơn bạn nhiều nha!!!