Tính
a)\(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)
b)\(\frac{18}{2.5}+\frac{18}{5.8}+...+\frac{18}{203.206}\)
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
9 tháng 10 2018
Bài 1:
\(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)
\(=2\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=2.\frac{32}{99}=\frac{64}{99}\)
Bài 2:
a) \(2.4^x-18=110\)
\(\Leftrightarrow2.4^x=128\)
\(\Leftrightarrow4^x=64\)
\(\Leftrightarrow4^x=4^3\Leftrightarrow x=3\)
Vậy x = 3
b) \(\left(\frac{3}{2}x-1\right)^5=1\)
\(\Leftrightarrow\frac{3}{2}x-1=1\)
\(\Leftrightarrow\frac{3}{2}x=2\)
\(\Leftrightarrow x=\frac{4}{3}\)
Vậy \(x=\frac{4}{3}\)
SN
9 tháng 10 2018
a) 4/3.5 + 3/5.7 + .... + 4/97.99
= 4( 1/3.5 +1/5.7 + ... + 1/97.99 )
= 4 . 1/2 . 2 ( 1/3.5 +1/5.7 + ... + 1/97.99 )
= 4/2 ( 2/3.5 + 2/5.7 + .... + 2/97.99 )
= 2 ( 5-3/3.5 + 7-5/5.7 + ..... + 99-97/97.99 )
= 2 (5/3.5 - 3/3.5 + 7/5.7 - 5/5.7 + .... + 99/97.99 - 97/97.99 )
= 2 ( 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/97 - 1/99 )
= 2 ( 1/3 -1/99 )
= 2 (33/99 - 1/99 )
= 2 . 32/99
= 32.2/99
=64/99
PM
1
31 tháng 3 2018
Ta có :
\(\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{97.99}\)
\(=\)\(2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\right)\)
\(=\)\(2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\)\(2\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=\)\(2.\frac{32}{99}\)
\(=\)\(\frac{64}{99}\)
Chúc bạn học tốt ~
TV
1
20 tháng 2 2017
de y \(\frac{1}{3}\)-\(\frac{1}{5}\)=\(\frac{2}{3.5}\)
tuong tu suy ra
M=\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+......+\(\frac{1}{97}\)-\(\frac{1}{99}\)
M=\(\frac{1}{3}\)-\(\frac{1}{99}\)
M=\(\frac{32}{99}\)
NT
7 tháng 5 2018
Bài 1 :
Ta có :
\(A=\frac{10^{17}+1}{10^{18}+1}=\frac{\left(10^{17}+1\right).10}{\left(10^{18}+1\right).10}=\frac{10^{18}+10}{10^{19}+10}\)
Mà : \(\frac{10^{18}+10}{10^{19}+10}>\frac{10^{18}+1}{10^{19}+1}\)
Mà \(A=\frac{10^{18}+10}{10^{19}+10}\)nên \(A>B\)
Vậy \(A>B\)
Bài 2 :
Ta có :
\(S=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)
\(\Rightarrow S=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2013+3}{2013}\)
\(\Rightarrow S=1-\frac{1}{2014}+1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{3}{2013}\)
\(\Rightarrow S=4+\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)
Vì \(\frac{1}{2013}>\frac{1}{2014}>\frac{1}{2015}>\frac{1}{2016}\)nên \(\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)
Nên : \(M>4\)
Vậy \(M>4\)
Bài 3 :
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{100^2}\)
Suy ra : \(A< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+....+\frac{1}{99.101}\)
\(\Rightarrow A< \frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{99.101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-......-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+......+\frac{1}{101}\right)\right]\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}\right)\)
\(\Rightarrow A< \frac{3}{4}\)
Vậy \(A< \frac{3}{4}\)
Bài 4 :
\(a)A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{1}{2015.2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{2016}{2017}\)
\(\Rightarrow A=\frac{1008}{2017}\)
Vậy \(A=\frac{1008}{2017}\)
\(b)\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{x\left(x+2\right)}=\frac{1008}{2017}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{x.\left(x+2\right)}=\frac{2016}{2017}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2016}{2017}\)
\(1-\frac{1}{x+2}=\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{2017}\)
\(\Rightarrow x+2=2017\)
\(\Rightarrow x=2017-2=2015\)
Vậy \(x=2015\)
YS
4
27 tháng 1 2016
mk làm đc,vs điều kiện bạn phải tick cho mk khi mk giải xog, ko đc tick cho ai
25 tháng 4 2017
ta co : 65%=0,65
goi A= 4.(1/3.5+1/5.7+1/7.9+............+1/97.99)
2A=4.( 2/3.5+2/5.7+2/7.9+...............+2/97.99)
2A=4.(1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99)
2A=4.(1/3-1/99)
2A=4.(33/=99+1/99)
2A=4.34/99
2A=136/99
A=136/99:2
A=68/99=0,69=0,68
Vi A=0,68 > 0,65
=> A > 65%
11 tháng 4 2019
a) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(=5.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
\(=5.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right):2\)
\(=5.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right):2\)
\(=5.\left(1-\frac{1}{101}\right):2=5.\frac{100}{101}:2=\frac{500}{101}.\frac{1}{2}\)\(=\frac{250}{101}\)
b) \(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(=3\left(\frac{1}{3.6}+\frac{1}{6.9}+...+\frac{1}{30.33}\right)\)\(.\frac{1}{3}\)
\(=(\frac{3}{3.6}+\frac{3}{6.9}+...+\frac{3}{30.33}).\frac{1}{3}\)
\(=(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}).\frac{1}{3}\)
\(=(\frac{1}{3}-\frac{1}{33}).\frac{1}{3}=\frac{10}{33}.\frac{1}{3}=\frac{10}{99}\)
a) \(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)
\(=4.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\right)\)
\(=4.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=4.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=4.\frac{32}{99}\)
\(=\frac{128}{99}\)
\(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)
\(=2\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=2.\frac{32}{99}\)
\(=\frac{64}{99}\)