a) Tính nhanh: \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
b) Cho A = \(\frac{3}{4}\times x+7\) ; B = \(\frac{4}{3}\times x-35\)
Tìm x để A và B có giá trị bằng nhau?
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 5 2017
a) \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=\frac{1}{2}.\frac{10}{39}\)
\(=\frac{5}{39}\)
9 tháng 5 2017
a)1/3.5+1/5.7+...+1/11.13
=1/2x(1/3-1/5+1/5-1/7+...+1/11-1/13)
=1/2x(1/3-1/13)
=1/2x10/39
=5/39
U
25 tháng 7 2018
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{13\cdot15}\)
\(=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{13\cdot15}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{15}\right)\)
\(=\frac{1}{2}\cdot\frac{14}{15}\)
\(=\frac{7}{15}\)
25 tháng 7 2018
Sửa đề chút nhé:
\(\left(1+3+5+7+...+2009+2011\right).\left(125125.127-127127.125\right)\)
\(=\left(1+3+5+7+...+2009+2011\right).\left(125.1001.127-127.1001.125\right)\)
\(=\left(1+3+5+7+...+2009+2011\right).0\)
\(=0\)
Ý b tham khảo bài bạn nguyen thi thuy linh nhé
XO
18 tháng 8 2020
1) a.Ta có \(A=\frac{3n+9}{n-4}=\frac{3n-12+21}{n-4}=\frac{3\left(n-4\right)}{n-4}+\frac{21}{n-4}=3+\frac{21}{n-4}\)
Vì \(3\inℤ\Rightarrow\frac{21}{n-4}\inℤ\Rightarrow21⋮n-4\Rightarrow n-4\inƯ\left(21\right)\)
=> \(n-4\in\left\{1;-1;3;-3;7;-7;21;-21\right\}\)
=> \(n\in\left\{5;3;8;1;11;-3;25;-17\right\}\)
b) Ta có B = \(\frac{6n+5}{2n-1}=\frac{6n-3+8}{2n-1}=\frac{3\left(2n-1\right)+8}{2n-1}=3+\frac{8}{2n-1}\)
Vì \(3\inℤ\Rightarrow\frac{8}{2n-1}\inℤ\Rightarrow2n-1\inƯ\left(8\right)\Rightarrow2n-1\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)(1)
lại có với mọi n nguyên => 2n \(⋮\)2 => 2n - 1 không chia hết cho 2 (2)
Kết hợp (1) ; (2) => \(2n-1\in\left\{1;-1\right\}\Rightarrow n\in\left\{1;0\right\}\)
2) Ta có : \(\frac{5}{x}+\frac{y}{4}=\frac{1}{8}\)
=> \(\frac{20+xy}{4x}=\frac{1}{8}\)
=> 4x = 8(20 + xy)
=> x = 2(20 + xy)
=> x = 40 + 2xy
=> x - 2xy = 40
=> x(1 - 2y) = 40
Nhận thấy : với mọi y nguyên => 1 - 2y là số không chia hết cho 2 (1)
mà x(1 - 2y) = 40
=> 1 - 2y \(\inƯ\left(40\right)\)(2)
Kết hợp (1) (2) => \(1-2y\in\left\{1;5;-1;-5\right\}\)
Nếu 1 - 2y = 1 => x = 40
=> y = 0 ; x = 40
Nếu 1 - 2y = 5 => x = 8
=> y = -2 ; x = 8
Nếu 1 - 2y = -1 => x = -40
=> y = 1 ; y = - 40
Nếu 1 - 2y = -5 => x = -8
=> y = 3 ; x =-8
Vậy các cặp (x;y) thỏa mãn là : (40 ; 0) ; (8; - 2) ; (-40 ; 1) ; (-8 ; 3)
4) \(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}=\frac{-\frac{19}{60}.\frac{5}{19}}{\frac{21}{70}.\frac{-4}{3}}=\frac{-\frac{5}{60}}{\frac{2}{5}}=-\frac{5}{60}:\frac{2}{5}=-\frac{5}{24}\)
b) \(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(6,3.12-21.3,6\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{100}}\)
\(=\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).0}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}=0\)
c) \(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}}=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}\right)}\)
\(=\frac{1}{4}+\frac{3}{4}=1\)
14 tháng 7 2018
Các bạn nêu rõ cách làm từng bài giúp mình nhé! Thanks ^-^!
N
25 tháng 7 2018
a) ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 )
vì ( 125125 x 127 - 127127 x 125 ) =[125125 x (125+2)] - 127127 x 125 ) =>125125 x (125+2)=125.125125+125125.2=125125.125+250250=125125.125+125.2002=125.(125125+2002)=125.127127
=> ( 125125 x 127 - 127127 x 125 )=127127.125-127127.125=0
=> (1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 ) =0
25 tháng 7 2018
a) ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 )
= ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x 0
= 0
b, \(\frac{1}{3}\)+ \(\frac{1}{15}\)+ \(\frac{1}{35}\)+ \(\frac{1}{63}\)+ \(\frac{1}{99}\)+ \(\frac{1}{143}\)+ \(\frac{1}{195}\)
= \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+\(\frac{1}{7}\)- \(\frac{1}{9}\)+...........+\(\frac{1}{13}\)- \(\frac{1}{15}\)
= \(\frac{1}{3}\)- \(\frac{1}{15}\)
= \(\frac{4}{15}\)
PL
6 tháng 3 2020
1. A = 75(42004 + 42003 +...+ 42 + 4 + 1) + 25
A = 25 . [3 . (42004 + 42003 +...+ 42 + 4 + 1) + 1]
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 3 + 1)
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 4)
A = 25 . 4 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1)
A =100 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1) \(⋮\) 100
XO
24 tháng 7 2019
a) \(\frac{4}{11}-\frac{7}{15}+\frac{7}{11}-\frac{5}{15}\)
\(=\left(\frac{4}{11}+\frac{7}{11}\right)-\left(\frac{7}{15}+\frac{5}{15}\right)\)
\(=1-\frac{4}{5}\)
\(=\frac{1}{5}\)
b) \(\frac{7}{3}-\frac{4}{9}-\frac{1}{3}-\frac{5}{9}\)
\(=\left(\frac{7}{3}-\frac{1}{3}\right)-\left(\frac{4}{9}+\frac{5}{9}\right)\)
\(=2-1\)
\(=1\)
c) \(\frac{1}{4}+\frac{7}{33}-\frac{5}{3}\)
\(=\frac{-1}{4}+\frac{-16}{11}\)
\(=\frac{-75}{44}\)
d) \(\frac{-3}{4}\times\frac{8}{11}-\frac{3}{11}\times\frac{1}{2}\)
\(=\frac{-6}{11}-\frac{3}{22}\)
\(=\frac{15}{22}\)
e) \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+\frac{1}{11\times13}+\frac{1}{13\times15}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(=\frac{1}{3}-\frac{1}{15}\)
\(=\frac{4}{15}\)
PM
23 tháng 4 2018
2S=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\)
=\(1-\frac{1}{15}=\frac{14}{15}\)
\(\Rightarrow S=\frac{7}{15}\)
23 tháng 4 2018
a. Ta có:A= 1/1.3+1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15
A=1/2(1/1.3+1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15)
A=1/2(1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13+1/13-1/15)
A=2(1-1/15)
A=1/2.14/15
A=7/15
a)Ta có:
A= 1/15+1/35+1/63+1/99+1/143
A= 1/3.5+1/5.7+1/7.9+1/9.11+1/11.13
2A= 2/3.5+2/5.7+2/7.9+2/9.11+2/11.13
2A= 1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13
Đơn giản đi ta được:
2A= 1/3-1/13
2A= 10/39
A= 5/39
Vậy A= 5/39
b) Để A và B có giá trị bằng nhau thì:
\(\frac{3}{4}\cdot x+7=\frac{4}{3}\cdot x-35\)
\(7+35=\frac{4}{3}\cdot x-\frac{3}{4}\cdot x\)
\(42=\frac{7}{12}\cdot x\)
\(x=42:\frac{7}{12}\)
\(x=72\)