Chứng tỏ rằng :
\(\dfrac{7}{12}\)<\(\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+............+\dfrac{1}{40}\)<\(\dfrac{5}{6}\)
K
Khách
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Những câu hỏi liên quan
A
24 tháng 3 2021
Ta có:
A=9999931999−5555571997
A=9999931998.999993−5555571996.555557
A=(9999932)999.999993 − (5555572)998.555557
A=\(\overline{\left(....9\right)}^{999}\) . 999993 - \(\overline{\left(...1\right)}.\text{555557}\)
A=\(\overline{\left(...7\right)}-\overline{\left(...7\right)}\)
A= \(\overline{\left(...0\right)}\)
Vì A có tận cùng là 0 nên \(A⋮5\)
NA
Ngoc Anh Thai
Giáo viên
11 tháng 4 2021
a)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{29}-\dfrac{1}{30}\\ =1-\dfrac{1}{30}=\dfrac{29}{30}< 1\left(dpcm\right)\)
b)
\(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}=\dfrac{1}{10}+\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\\ >\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{1}{10}+\dfrac{90}{100}\\ =\dfrac{110}{100}>1\left(đpcm\right).\)
NA
Ngoc Anh Thai
Giáo viên
11 tháng 4 2021
c)
\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\\ =\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}\right)\\ < \dfrac{1}{5}.5+\dfrac{1}{8}.8=1+1=2\left(đpcm\right)\)
d) tương tự câu 1
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 3 2022
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{20}-\dfrac{1}{20}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{20}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{10}\right)\)
\(=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\) (đpcm)
AH
Akai Haruma
Giáo viên
23 tháng 3 2023
Lời giải:
$P< \frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{199.201}+\frac{1}{201.203}$
$P< \frac{1}{2}(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{199.201}+\frac{2}{201.203})$
$P< \frac{1}{2}(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{199}-\frac{1}{201}+\frac{1}{201}-\frac{1}{203})$
$P< \frac{1}{2}(\frac{1}{3}-\frac{1}{203})< \frac{1}{2}.\frac{1}{3}=\frac{1}{6}$
DT
Đỗ Thanh Hải
CTVVIP
2 tháng 5 2021
Ta có
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
...............
\(\dfrac{1}{8^2}< \dfrac{1}{7.8}\)
=> B < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{7.8}\)
B < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
B < \(1-\dfrac{1}{8}< 1\) (Do \(\dfrac{1}{8}>0\))
Vậy.....
QT
27 tháng 6 2018
Gọi biểu thức là \(A\). Ta có :
\(A=\dfrac{3}{1.2.3}+\dfrac{5}{2.3.4}+\dfrac{7}{3.4.5}+...+\dfrac{2017}{1008.1009.1010}\)
\(A=\left(\dfrac{1.2}{1.2.3}+\dfrac{2.2}{2.3.4}+\dfrac{3.2}{3.4.5}+...+\dfrac{1008.2}{1008.1009.1010}\right)+\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{1008.1009.1010}\right)\)\(A=\left(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{1009.1010}\right)+\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{1008.1009}-\dfrac{1}{1009.1010}\right)\)
\(A=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{1009}-\dfrac{1}{1010}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{1009.1010}\right)\)
\(A< 2.\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{1}{2}=1+\dfrac{1}{4}=\dfrac{5}{4}\)
SG
4
VV
15 tháng 5 2017
Ta có :\(\dfrac{1}{5}< \dfrac{1}{4};\dfrac{1}{6}< \dfrac{1}{4};\dfrac{1}{7}< \dfrac{1}{4};\dfrac{1}{8}< \dfrac{1}{4}\)
\(\Rightarrow\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{8}< \dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{4}{4}=1\left(1\right)\)
Ta có :\(\dfrac{1}{9}< \dfrac{1}{8};\dfrac{1}{10}< \dfrac{1}{8};\dfrac{1}{11}< \dfrac{1}{8};...;\dfrac{1}{17}< \dfrac{1}{8}\)
\(\Rightarrow\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}< \dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+...+\dfrac{1}{8}=\dfrac{8}{8}=1\left(2\right)\)
Từ (1) và (2)\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 1+1=2\)
Vậy \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)
14 tháng 3 2018
Ta có : \(\dfrac{1}{5}=\dfrac{1}{5}\), \(\dfrac{1}{6}< \dfrac{1}{5}\), \(\dfrac{1}{7}< \dfrac{1}{5}\),...,\(\dfrac{1}{9}< \dfrac{1}{5}\)
Vậy \(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}< \dfrac{1}{5}\cdot5=1\)
\(\dfrac{1}{10}< \dfrac{1}{8},\dfrac{1}{11}< \dfrac{1}{8},...,\dfrac{1}{17}< \dfrac{1}{8}\)
Vậy \(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}< \dfrac{1}{8}\cdot8=1\)
Vậy \(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{17}< 1+1=2\)
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