Cho A = 1/6*25+1/7*30+1/8*35+...+1/100*495. Chứng tỏ rằng A <1/25
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Những câu hỏi liên quan
25 tháng 10 2023
Bài 1:vì 15 chia hết cho 5 suy ra 2022.15 chia hết cho 5
vì 25 chia hết cho 5 suy ra 2022.15 + 25 chia hết cho 5
TV
2
NN
6
9 tháng 8 2016
Ta có:
\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}< \frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\)
Mà \(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{1}{4}.4=1\)
=>\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}< 1\) (1)
\(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\)Mà \(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{8}.8=1\)
=> \(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}< 1\) (2)
Từ (1) và (2)
=> A=\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+...+\frac{1}{14}+\frac{1}{15}< 1+1\)
=> A<2
NP
0
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
Ta có:
\(A=\frac{1}{6.25}+\frac{1}{7.30}+...+\frac{1}{8.35}+\frac{1}{100.495}\)
\(=\frac{1}{6.\left(5.5\right)}+\frac{1}{7.\left(5.6\right)}+...+\frac{1}{8.\left(5.7\right)}+\frac{1}{100.\left(5.99\right)}\)
\(=\frac{1}{5}\left(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{99.100}\right)\)
\(=\frac{1}{5}\left[\left(\frac{1}{5}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{8}\right)+...+\left(\frac{1}{99}-\frac{1}{100}\right)\right]\)
\(=\frac{1}{5}\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{1}{5}\left(\frac{1}{5}-\frac{1}{100}\right)\)
Mà \(\frac{1}{5}-\frac{1}{100}< \frac{1}{5}\)nên \(A=\frac{1}{5}\left(\frac{1}{5}-\frac{1}{100}\right)< \frac{1}{5}.\frac{1}{5}=\frac{1}{25}.\)
Vậy \(A< \frac{1}{25}.\)
100-5=95 phân số
(1/100+1/6):2=53/600
(495-25):5+1=95 số
(495+5)x95:2=23750
53/600x23750=25175/12