A=1+1+1+...+1-99
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
DV
0
TL
1
18 tháng 4 2016
A = 99^2015 + 1/99^2014 + 1 < 99^2015 + 1 + 98 / 99^2014 + 1 + 98
= 99^2015 + 99 / 99^2014 + 99
= 99(99^2014 + 1) / 99(99^2013+1)
= 99^2014 + 1 / 99^2013 + 1 = B
=> A < B
P
1
25 tháng 8 2023
a: \(A=\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)+1\)
\(=\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}+\dfrac{100}{100}\)
\(=100\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)\)=100B
=>B/A=1/100
b: \(A=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\left(1\right)\)
\(=\dfrac{50}{49}+\dfrac{50}{48}+....+\dfrac{50}{2}+\dfrac{50}{50}\)
\(=50\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)
\(B=\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{49}+\dfrac{2}{50}\)
\(=2\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)\)
=>A/B=25
10 tháng 8 2015
\(\Rightarrow\) Tử = (\(\frac{1}{99}\)+1)+(\(\frac{2}{98}\)+1)+ \(\left(\frac{3}{97}+1\right)\)+.....+ \(\left(\frac{97}{3}+1\right)\)+ \(\left(\frac{98}{2}+1\right)\) + 1
= \(\frac{100}{99}\) + \(\frac{100}{98}\) + \(\frac{100}{97}\) + ......\(+\frac{100}{3}\).+ \(\frac{100}{2}\) + \(\frac{100}{100}\)
= 100\(\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+....+\frac{1}{3}+\frac{1}{2}+\frac{1}{100}\right)\)
= 100\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}+\frac{1}{100}\right)\) giá trị trong dấu ngoặc bằng mẫu số .
\(\Rightarrow\)Nên kết quả là 100
21 tháng 6 2018
\(\Rightarrow A=\frac{1}{99.100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\right)\)
\(\Leftrightarrow A=\frac{1}{99.100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(\Rightarrow A=\frac{1}{99.100}-1+\frac{1}{99}\)
24 tháng 3 2018
\(A=\frac{\frac{1}{99}+\frac{1}{98}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)
\(A=\frac{\left(\frac{1}{99}+1\right)+\left(\frac{1}{98}+1\right)+...+\left(\frac{98}{2}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)
\(A=\frac{\frac{100}{99}+\frac{100}{98}+...+\frac{100}{2}+\frac{100}{100}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)
\(A=\frac{100\left(\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}+\frac{1}{100}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)
\(A=100\)
DT
0
How many one are there ?