Tính:
a) \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)
b) \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\)
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
10 tháng 5 2017
\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+.....+\frac{3^2}{97.100}\)
\(=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\right)\)
Ta thấy :
\(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)
\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)
\(.........\)
\(\frac{3}{97.100}=\frac{100-97}{97.100}=\frac{1}{97}-\frac{1}{100}\)
\(\Rightarrow A=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{100}\right)\)
\(=3\left(1-\frac{1}{100}\right)=3\cdot\frac{99}{100}=\frac{297}{100}\)
10 tháng 5 2017
đáp án = \(\frac{297}{100}\)
đúng không?
kết bạn với mh nha
LM
5 tháng 8 2017
\(a,=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-0-0-0-...-0-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}\)
\(=\frac{4}{8}-\frac{1}{8}\)
\(=\frac{3}{8}\)
\(b,=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{49}+\frac{1}{49}-\frac{1}{16}\)
\(=1-0-0-0-...-0-\frac{1}{16}\)
\(=1-\frac{1}{16}\)
\(=\frac{15}{16}\)
\(c,\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\left(1-0-0-0-...-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\frac{50}{51}\)
\(=\frac{25}{17}\)
\(d,\)giống câu a tự làm nha mỏi tay quá.
BT
5 tháng 8 2017
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}.\)
=> \(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)
=> \(A=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
\(B=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{49.52}=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{49}-\frac{1}{52}\)
=> \(B=\frac{1}{4}-\frac{1}{52}=\frac{24}{104}=\frac{1}{26}\)
2 tháng 5 2017
A:3=\(\frac{3}{1.4}+\frac{3}{4.7}\)\(+.....+\frac{3}{97.100}\)
A:3=\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{100}\)
A:3=\(\frac{1}{1}-\frac{1}{100}\)
A:3=\(\frac{99}{100}\)
A=\(\frac{99}{100}.3\)
A=\(\frac{297}{100}\)
BB
2 tháng 5 2017
\(A:3=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\)
\(A:3=\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A:3=\frac{1}{1}-\frac{1}{100}\)
\(A:3=\frac{99}{100}\)
\(A=\frac{99}{100}.3\)
\(A=\frac{297}{100}\)
TD
1
25 tháng 2 2020
B =\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...\frac{2}{97.100}\)
=2.(\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\))
3B=2.(\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\))
3B=2.(\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\))
3B=2.(1-\(\frac{1}{100}\))
3B=2.\(\frac{99}{100}\)=\(\frac{99}{50}\)
=>B=\(\frac{99}{50}:3\)=\(\frac{33}{50}\)
Tick mik nha
DN
1
4 tháng 8 2016
\(B=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+..+\frac{3}{97.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-....+\frac{1}{94}-\frac{1}{97}+\frac{1}{97}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}\)
\(=\frac{100}{100}-\frac{1}{100}\)
\(=\frac{99}{100}\)
24 tháng 3 2019
a, \(\frac{9}{1.2}+\frac{9}{2.3}+...+\frac{9}{99.100}\)
=9.(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\))
= 9(1 -\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\))
=9(1-\(\frac{1}{100}\))
A=\(\frac{891}{100}\)
b, \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{27.30}\)
=1-(\(\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{27}-\frac{1}{30}\))
=1-\(\frac{1}{30}\)
B=\(\frac{29}{30}\)
24 tháng 3 2019
a) \(\dfrac{9}{1.2}+\dfrac{9}{2.3}+...+\dfrac{9}{99.100}\)
\(=9\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)
\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=9\left(1-\dfrac{1}{100}\right)\)
\(=9.\dfrac{99}{100}\)
\(=\dfrac{891}{100}\)
b) \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{27.30}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{27}-\dfrac{1}{30}\)
\(=1-\dfrac{1}{30}\)
\(=\dfrac{29}{30}\)
19 tháng 5 2017
\(\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\frac{3^2}{10.13}+...+\frac{3^2}{97.100}\)
\(=3.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{97.100}\right)\)
\(=3.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=3.\left(1-\frac{1}{100}\right)\)
\(=3.\frac{99}{100}\)
T
25 tháng 3 2018
Bài nhìn vô muốn xỉu rồi ='((
1. a) \(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{91.94}+\frac{2}{94.97}\)
\(=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{91.94}+\frac{3}{94.97}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{97}\right)=\frac{2}{3}.\frac{96}{97}=\frac{64}{97}\)
b) Bạn tự làm, làm nữa chắc xỉu =((( Khi nào rảnh mình sẽ làm, nếu bạn cần
2 )
a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{1005}{2011}\)
\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{1005}{2011}:2=\frac{1005}{4022}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{1005}{4022}=\frac{3017}{4020+2}\)
\(\Rightarrow x=4020\)
a) 1/1 - 1/3 +1/3 - 1/5 +........+1/49 - 1/51
=1/1 - 1/51 (các số liền kề nhau cộng lại bằng 0)
=50/51
còn câu b bạn tự giải
nhớ thank mik nha!!!!!
b,khoảng cách của nó là 3 mà tử của nó bằng 3 chứng tỏ nó là dạng đủ
1/1-1/4+1/4-1/7+...+1/97-1/100
1-1/100=99/100