TX
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
CY
7
2 tháng 4 2019
\(A=4\left(\frac{1}{20}-\frac{1}{80}\right)=4.\frac{3}{80}=60\)
2 tháng 4 2019
\(A=\frac{16}{20\cdot24}+\frac{16}{24\cdot28}+\frac{16}{28\cdot32}+...+\frac{16}{76\cdot80}\)
\(A=4\left[\frac{4}{20\cdot24}+\frac{4}{24\cdot28}+\frac{4}{28\cdot32}+...+\frac{4}{76\cdot80}\right]\)
\(A=4\left[\frac{1}{20}-\frac{1}{24}+...+\frac{1}{76}-\frac{1}{80}\right]\)
\(A=4\left[\frac{1}{20}-\frac{1}{80}\right]\)
\(A=4\left[\frac{4}{80}-\frac{1}{80}\right]=4\cdot\frac{3}{80}=\frac{4\cdot3}{80}=\frac{1\cdot3}{20}=\frac{3}{20}\)
23 tháng 2 2020
Ta có:\(\frac{1}{2^2}=\frac{1}{4};\frac{1}{3^2}< \frac{1}{2\cdot3}=\frac{1}{2}-\frac{1}{3};\frac{1}{3^2}< \frac{1}{3\cdot4}=\frac{1}{3}-\frac{1}{4};.....;\frac{1}{100^2}< \frac{1}{99\cdot100}=\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{4}+\frac{1}{2}-\frac{1}{100}< \frac{3}{4}\left(đpcm\right)\)
23 tháng 2 2020
Gọi \(D=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}+\frac{1}{100^2}< \frac{3}{4}\)
Vì \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{100^2}< \frac{1}{99.100}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}< \frac{3}{4}\)
\(\Rightarrow D< \frac{3}{4}\left(đpcm\right)\)
HD
0
VH
1
DY
13 tháng 3 2019
ta có : 1/2^2<1/2x3
1/3^2<1/3x4
...........
1/100^2<1/99x100
suy ra :1/2^2+1/3^2 +........+1/100^2<1/2x3+1/3x4+1/4x5+..........+1/99x100
Gọi A=1/2x3+1/3x4+............+1/99x100
A=3-2/2x3+4-3/3x4+..........+100-99/99x100
A=3/2x3-2/2x3+4/3x4-3/3x4+........+100/99x100-99/99x100
A=1/2-1/100
A=49/100
1/2^2+1/3^2+......+1/100^2<49/100
Ta có:3/4=75/10049/100
Mà 75/100>49/100
1/2^2+1/3^2+........+1/100^2<3/4
NV
0
6 tháng 3 2019
\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\)
\(=\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)< \frac{1}{4}\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\right)\)
\(=\frac{1}{4}\left(1-\frac{1}{n}\right)\)(đpcm)
6 tháng 3 2019
Ta có:\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)
\(=\frac{1}{4.4}+\frac{1}{4.9}+\frac{1}{4.16}+...+\frac{1}{4.n^2}\)
\(=\frac{1}{4}\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{n^2}\right)\)
\(Xét:\)
\(\frac{1}{2.2}< \frac{1}{1.2};\frac{1}{3.3}< \frac{1}{2.3};\frac{1}{4.4}< \frac{1}{3.4};\frac{1}{n.n}< \frac{1}{\left(n-1\right).n}...\)
\(Suyra:\)
\(P=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{n.n}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)
\(\Leftrightarrow P< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(\Leftrightarrow P< 1-\frac{1}{n}< 1\)
\(\Leftrightarrow\frac{1}{4}.P< 1.\frac{1}{4}\)
\(\Leftrightarrow\frac{1}{4}\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{n^2}\right)< \frac{1}{4}\)
\(\Leftrightarrow\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\left(đpcm\right)\)
DT
0
\(=4\left(\frac{4}{20.24}+\frac{4}{24.28}+...+\frac{4}{76.80}\right)\)
\(=4\left(\frac{1}{20}-\frac{1}{24}+\frac{1}{24}-\frac{1}{28}+...+\frac{1}{76}-\frac{1}{80}\right)\)
\(=4\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=4\times\frac{3}{80}\)
\(=\frac{12}{80}=\frac{3}{20}<1\)
đpcm
frtji