A = 3/4.8/9.15/16....2499/2500
B = 3/3.5 + 3/5.7 + 3/7.9 + .....3/47.49
giúp em với câu này khó quá
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 3 2016
B = 3/4.8/9.15/16.....2499/2500
= 1.3/2^2 . 2.4/3^2 . 3.5/4^2 ..... 49.51/50^2
= 1.2.3.....49/1.2.3.....50 . 3.4.5....51/1.2.3.....50
= 1/50 . 51/2
= 51/100
CX
24 tháng 3 2016
Giải
B = 3/4.8/9.15/16.....2499/2500
B=1.3/2^2 . 2.4/3^2 . 3.5/4^2 ..... 49.51/50^2
B=1.2.3.....49/2.3.4.....50 . 3.4.5.....51/2.3.4.....50
B=1/50 . 51/2
B=51/100
AH
Akai Haruma
Giáo viên
14 tháng 9
Lời giải:
\(A=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}....\frac{-998}{999}.\frac{-999}{1000}\\
=\frac{(-1)(-2)(-3)...(-998)(-999)}{2.3.4....1000}\\
=-\frac{1.2.3.4....998.999}{2.3.4...1000}\\
=-\frac{1}{1000}\)
AH
Akai Haruma
Giáo viên
14 tháng 9
Trong $B$ có một thừa số là $1-\frac{7}{7}=0$ nên $B=0$ (do số nào nhân với $0$ cũng sẽ bằng $0$.
----------------------
$C=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{49.51}{50^2}$
$=\frac{1.3.2.4.3.5.....49.51}{2^2.3^2.4^2....50^2}$
$=\frac{(1.2.3...49)(3.4.5...51)}{(2.3.4...50)(2.3.4...50)}$
$=\frac{1.2.3...49}{2.3.4...50}.\frac{3.4.5...51}{2.3.4....50}$
$=\frac{1}{50}.\frac{51}{2}=\frac{51}{100}$
H
15 tháng 4 2019
\(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)
\(=\frac{1}{3}-\frac{1}{21}\)
\(=\frac{7}{21}-\frac{1}{21}=\frac{6}{21}\)
15 tháng 4 2019
\(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)
\(A=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+\left(\frac{1}{9}-\frac{1}{9}\right)+...+\left(\frac{1}{19}-\frac{1}{19}\right)-\frac{1}{21}\)
\(A=\frac{1}{3}-\frac{1}{21}\)
\(A=\frac{2}{7}\)
GM
1
3 tháng 4 2017
a.
\(M=1.\left[\frac{1}{3}-\frac{1}{5}+.....\frac{1}{97}-\frac{1}{99}\right]\)
\(M=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
b.
\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right]\)
\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{199}\right]=\frac{291}{995}\)
mk đầu tiên nha bạn
17 tháng 12 2023
\(\dfrac{2^3}{3\cdot5}+\dfrac{2^3}{5\cdot7}+...+\dfrac{2^3}{101\cdot103}\)
\(=2^2\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{101\cdot103}\right)\)
\(=4\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{101}-\dfrac{1}{103}\right)\)
\(=4\cdot\left(\dfrac{1}{3}-\dfrac{1}{103}\right)\)
\(=4\cdot\dfrac{100}{309}=\dfrac{400}{309}\)
7 tháng 7 2017
G=\(\frac{3}{2.5}+\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{2015.2017}\)
G=\(3.\left(\frac{1}{2.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}\right)\)
G=\(3.\left(\frac{1}{2}.\frac{1}{5}+\frac{1}{5}.\frac{1}{7}+\frac{1}{7}.\frac{1}{9}+...+\frac{1}{2013}.\frac{1}{2015}+\frac{1}{2015}.\frac{1}{2017}\right)\)
G=\(3.\left(\frac{1}{2}+\frac{1}{2017}\right)\)
G=1.5
Anh ko bik có đúng ko nữa lâu quá rồi. Em thông cảm nhé
BC
31 tháng 3 2019
a)Ta có:
\(A=4\frac{25}{16}+25\left(\frac{9}{16}:\frac{125}{64}\right):\frac{-27}{8}\)
\(\Rightarrow A=\frac{89}{16}+25.\frac{36}{125}:\frac{-27}{8}\)
\(\Rightarrow A=\frac{89}{16}+\frac{36}{5}:\frac{-27}{8}\)
\(\Rightarrow A=\frac{89}{16}+\frac{-32}{15}\)
\(\Rightarrow A=\frac{823}{240}\)
Vậy A=.....
b)Ta có:
\(C=\frac{2^3}{3.5}+\frac{2^3}{5.7}+\frac{2^3}{7.9}+...+\frac{2^3}{101.103}\)
\(\Rightarrow C=\frac{2^2.2}{3.5}+\frac{2^2.2}{5.7}+\frac{2^2.2}{7.9}+...+\frac{2^2.2}{101.103}\)
\(\Rightarrow C=2^2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{101.103}\right)\)
\(\Rightarrow C=4\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{101}-\frac{1}{103}\right)\)
\(\Rightarrow C=4\left(\frac{1}{3}-\frac{1}{103}\right)\)
\(\Rightarrow C=4.\frac{100}{309}\)
\(\Rightarrow C=\frac{400}{309}\)
Vậy C=.....
3 tháng 4 2017
\(A=\dfrac{3}{3.5} + \dfrac{3}{5.7} + ... + \dfrac{3}{97.99}\)
\(\Rightarrow A=\dfrac{3}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right)\)
\(\Rightarrow A=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(\Rightarrow A=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
\(\Rightarrow A=\dfrac{3}{2}.\dfrac{32}{99}\)
\(\Rightarrow A=\dfrac{16}{33}\)
Vậy \(A=\dfrac{16}{33}\)
DA
3 tháng 4 2017
A= \(\dfrac{3}{3.5}+\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{97.99}\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
= \(\dfrac{3}{2}.\dfrac{32}{99}\)
= \(\dfrac{3.32}{2.99}\)= \(\dfrac{3.2.3.6}{2.11.3.3}\)= \(\dfrac{6}{11}\)
B=\(\dfrac{3}{3.5}.\dfrac{3}{5.7}.....\dfrac{3}{47.49}\)
B=\(\dfrac{3}{2}.\left(\dfrac{2}{3.5}.\dfrac{2}{5.7}.....\dfrac{2}{47.49}\right)\)
B=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{47}-\dfrac{1}{49}\right)\)
B=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)
B=\(\dfrac{3}{2}.\dfrac{46}{147}\)
B=\(\dfrac{23}{49}\)
a) Ta có: \(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{2499}{2500}\)
\(=\dfrac{1\cdot3}{2^2}\cdot\dfrac{2\cdot4}{3^2}\cdot\dfrac{3\cdot5}{4^2}\cdot...\cdot\dfrac{49\cdot51}{50^2}\)
\(=\dfrac{1}{50}\cdot\dfrac{51}{2}=\dfrac{51}{100}\)
b) Ta có: \(B=\dfrac{3}{3\cdot5}+\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+...+\dfrac{3}{47\cdot49}\)
\(=\dfrac{3}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{47\cdot49}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{46}{147}=\dfrac{138}{294}=\dfrac{23}{49}\)