Không dùng máy tính bỏ túi hãy chứng minh S >1
S = \(\dfrac{5}{20}\)+\(\dfrac{5}{21}\)+\(\dfrac{5}{22}\)+\(\dfrac{5}{23}\)+\(\dfrac{5}{24}\)
Không dùng máy tính bỏ túi hãy chứng minh S >1
S = \(\dfrac{5}{20}\)+\(\dfrac{5}{21}\)+\(\dfrac{5}{22}\)+\(\dfrac{5}{23}\)+\(\dfrac{5}{24}\)
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
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
Ta có :
\(\frac{5}{20}>\frac{5}{25};\frac{5}{21}>\frac{5}{25};\frac{5}{22}>\frac{5}{25};\frac{5}{23}>\frac{5}{25};\frac{5}{24}>\frac{5}{25}\)
\(=>S< \frac{5}{25}+\frac{5}{25}+\frac{5}{25}+\frac{5}{25}+\frac{5}{25}\)
Mà : \(\frac{5}{25}+\frac{5}{25}+\frac{5}{25}+\frac{5}{25}+\frac{5}{25}=1\)
\(=>S< 1\)
Vậy bài toán đã được chứng minh.
Chứng minh rằng: \(\dfrac{1 + 3 + 5 + ... + 39}{21 + 22 + 23 + ... + 40} = \dfrac{1}{2^{20}}\)
Trong các phân số \(\dfrac{5}{4}\);\(\dfrac{22}{23}\);\(\dfrac{9}{9}\);\(\dfrac{24}{23}\) phân số bé nhất là
A.\(\dfrac{24}{23}\) B.\(\dfrac{22}{23}\) C.\(\dfrac{5}{4}\) D.\(\dfrac{9}{9}\)
`=>` `B`
Vì các phân số cồn lại thuộc dạng `(x >= 1)`
Chứng minh: \(\dfrac{1}{21}\)+\(\dfrac{1}{22}\)+\(\dfrac{1}{23}\)+\(\dfrac{1}{24}\)+....+\(\dfrac{1}{80}\)không phải là một số tự nhiên.
Tính giá trị biểu thức:
D=\(\left(\dfrac{136}{15}-\dfrac{28}{5}+\dfrac{62}{10}\right).\dfrac{21}{24}\)
F= \(\dfrac{5}{6}+6\dfrac{5}{6}.\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)
\(F=\dfrac{5}{6}+6\dfrac{5}{6}\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)
\(F=\dfrac{5}{6}+\dfrac{41}{6}\left(\dfrac{225}{20}-\dfrac{37}{4}\right):\dfrac{25}{3}\)
\(F=\dfrac{5}{6}+\dfrac{41}{6}.2.\dfrac{3}{25}\)
\(F=\dfrac{5}{6}+\dfrac{41}{25}.\dfrac{3}{25}\)
\(F=\dfrac{5}{6}+\dfrac{41}{25}\)
\(F=\dfrac{371}{150}\)
\(D=\left(\dfrac{136}{15}-\dfrac{28}{5}+\dfrac{62}{10}\right)\times\dfrac{21}{24}\)
\(D=\left(\dfrac{272}{30}-\dfrac{168}{30}+\dfrac{186}{30}\right)\times\dfrac{21}{24}\)
\(D=\dfrac{290}{30}\times\dfrac{21}{24}\)
\(D=\dfrac{29}{3}\times\dfrac{7}{8}\)
\(D=\dfrac{203}{24}\)
bài 1 : so sánh :
a) \(\dfrac{23}{21}\)và\(\dfrac{21}{23}\)
b)\(\dfrac{19}{26}\)và \(\dfrac{21}{25}\)
bài 2 : sắp sếp các phân số sau từ bé đến lớn :
a)\(\dfrac{7}{36};\dfrac{24}{36};\dfrac{13}{36};\dfrac{1}{36};\dfrac{43}{36};\dfrac{36}{36}\)
b)\(\dfrac{-3}{10};\dfrac{-31}{100};\dfrac{-297}{1000};\dfrac{10000}{-3056}\)
c)\(\dfrac{13}{20};\dfrac{7}{20};\dfrac{9}{4};\dfrac{2}{5};\dfrac{1}{2}\)
d)\(\dfrac{13}{21};\dfrac{152}{17};\dfrac{13}{17};\dfrac{-5}{21}\)
e)\(\dfrac{-1}{2};\dfrac{3}{-4};\dfrac{-2}{3};\dfrac{4}{-5}\)
\(S=\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}\) Chứng minh \(\dfrac{3}{5}< S< \dfrac{4}{5}\)
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
Cho tổng S= \(\dfrac{1}{31}\) + \(\dfrac{1}{32}\) + ... + \(\dfrac{1}{60}\) Chứng minh \(\dfrac{3}{5}\) < S < \(\dfrac{4}{5}\)
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
Câu 1: Tính giá trị biểu thức:
a.A=\(\left(\dfrac{136}{15}-\dfrac{28}{5}+\dfrac{62}{10}\right)\).\(\dfrac{21}{24}\)
b.B=\(\dfrac{5}{6}\)+6\(\dfrac{5}{6}\)\(\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right)\):8\(\dfrac{1}{3}\)
c.C=1+3+6+10+15+...+1225.