A = \(\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\)\(\frac{9}{99.100}\)
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
16 tháng 4 2018
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(A=9.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9.\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(A=9.\frac{99}{100}\)
\(A=\frac{891}{100}\)
8 tháng 8 2016
A=9.(1/1.2+1/2.3+1/3.4+....+1/98.99+1/99.100)
A=9.(1/1-1/2+1/2-1/3+...+1/98-1/99+1/99-1/100)
A=9.(1-1/100)
A=9.99/100
A=901/100
TN
23 tháng 4 2016
\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=9\left(1-\frac{1}{100}\right)\)
\(=9\times\frac{99}{100}\)
\(=\frac{891}{100}\)
23 tháng 4 2016
A=9.(1/1.2 +1/2.3 +1/3.4+...+1/98.99 +1/99.100
A=9.(1-1/2+1/2-1/3+1/3-1/4+...+1/98-1/99+1/99-1/100)
A=9.(1-1/100)
A=9.99/100
A=891/100
25 tháng 3 2015
Ta có:
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...\frac{9}{98.99}+\frac{9}{99.100}\)
\(=9.\frac{1}{1.2}+9.\frac{1}{2.3}+9.\frac{1}{3.4}+...+9.\frac{1}{98.99}+9.\frac{1}{99.100}\)
\(=9.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=9.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=9.\left(1-\frac{1}{100}\right)\)
\(=9.\frac{99}{100}\)
\(=\frac{9.99}{100}\)
\(=\frac{891}{100}\)
14 tháng 2 2016
=9.(1/1.2+1/2.3+1/3.4+...+1/98.99+1/99.100)
=9.(1/1-1/2+1/2-1/3+1/3-1/4+....+1/98-1/99+1/99-1/100)
=9.(1/1-1/100)
=9-9/100
=891/100
8 tháng 7 2017
\(\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(=9.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=9.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=9.\left(1-\frac{1}{100}\right)\)
\(=\frac{891}{100}\)
TM
8 tháng 7 2017
\(\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{99.100}=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=9\left(1-\frac{1}{100}\right)=9.\frac{99}{100}=\frac{891}{100}\)
29 tháng 3 2018
\(\frac{9}{1.2}+\frac{9}{2.3}+....+\frac{9}{98.99}+\frac{9}{99.100}\)
\(=9.\frac{1}{1.2}+9.\frac{1}{2.3}+....+9.\frac{1}{98.99}+9.\frac{1}{99.100}\)
\(=9.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=9.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=9.\left(1-\frac{1}{100}\right)=9.\frac{99}{100}=\frac{891}{100}\)
\(A=9.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9.\left(1+\left[-\frac{1}{2}+\frac{1}{2}\right]+\left[-\frac{1}{3}+\frac{1}{3}\right]+...+\left[-\frac{1}{99}+\frac{1}{99}\right]-\frac{1}{100}\right)\)
\(A=9.\left(1+0+0+...+0-\frac{1}{100}\right)\)
\(A=9.\left(1-\frac{1}{100}\right)\)
\(A=9.\left(\frac{100}{100}-\frac{1}{100}\right)=9.\left(\frac{99}{100}\right)\)
\(A=\frac{891}{100}=8\frac{91}{100}\)
k cho mk nha
\(A=\frac{9}{1.2}+\frac{9}{2.3}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=\frac{9.1}{1.2.1}+\frac{9.1}{2.3.1}+...+\frac{9.1}{98.99.1}+\frac{9.1}{99.100.1}\)
\(A=1\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(A=1\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=1\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(A=1.\frac{99}{100}\)
\(A=\frac{99}{100}\)