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NA
3
15 tháng 6 2017
Chuyển vế tất cả số hạng tự do sang phải, ta được \(x=1931\)bạn nhé!
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 3 2023
( \(\dfrac{2}{15}\) + \(\dfrac{2}{35}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
( \(\dfrac{2\times7}{15\times7}\) + \(\dfrac{2\times3}{35\times3}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
(\(\dfrac{14}{105}\) + \(\dfrac{6}{105}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
(\(\dfrac{20}{105}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
( \(\dfrac{4}{21}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
(\(\dfrac{12}{63}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
\(\dfrac{2}{9}\) : \(x\) = \(\dfrac{1}{18}\)
\(x\) = \(\dfrac{2}{9}\) : \(\dfrac{1}{18}\)
\(x\) = 4
DT
1
14 tháng 7 2018
Các bạn nêu rõ cách làm từng bài giúp mình nhé! Thanks ^-^!
NN
18 tháng 8 2017
<=> \(\left(\frac{1}{3\cdot5}+\frac{1}{5.7}+...+\frac{1}{13\cdot15}\right)+x=\frac{17}{15}\)
<=> \(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{15}\right)+x=\frac{17}{15}\)
<=>\(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{15}\right)+x=\frac{17}{15}\)
<=> \(\frac{2}{15}+x=\frac{17}{15}\)
=> x = 1
NH
18 tháng 8 2017
(1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15)+x=17/15
[2.(1/3-1/5+1/5-1/7+...+1/13-1/15)]+x=17/15
[2.(1/3-1/15)]+x=17/15
(2.4/15)+x=17/15
6/15+x=17/15
x=17/15-6/15
x=11/15
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
13 tháng 1 2024
a; (5142 - 17 x 8 + 242 : 11) x (27 - 3 x 9)
= (5142 - 17 x 8 + 242 : 11) x (27 - 27)
= (5142 - 17 x 8 + 242 : 11) x 0
= 0
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
13 tháng 1 2024
b;
(1 + \(\dfrac{1}{2}\)) \(\times\) (1 + \(\dfrac{1}{3}\)) \(\times\) ( 1 + \(\dfrac{1}{4}\)) \(\times\) ... \(\times\) (1 + \(\dfrac{1}{2010}\)) \(\times\)(1 + \(\dfrac{1}{2011}\))
= \(\dfrac{2+1}{2}\) \(\times\) \(\dfrac{3+1}{3}\) \(\times\) \(\dfrac{4+1}{4}\)\(\times\) ... \(\times\) \(\dfrac{2010+1}{2010}\)\(\times\) \(\dfrac{2011+1}{2011}\)
= \(\dfrac{3}{2}\)\(\times\)\(\dfrac{4}{3}\)\(\times\)\(\dfrac{5}{4}\)\(\times\)...\(\times\)\(\dfrac{2011}{2010}\)\(\times\)\(\dfrac{2012}{2011}\)
= \(\dfrac{2012}{2}\)
= 1006
22 tháng 6 2017
a)\(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\)
\(=\frac{13}{3.5}+\frac{13}{5.7}+\frac{13}{7.9}+\frac{13}{9.11}\)
\(=\frac{13}{2}\left(\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{13}{2}\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{13}{2}\cdot\frac{8}{33}\)
\(=\frac{52}{33}\)
22 tháng 6 2017
a) Đặt A= 13/15 + 13/35 + 13/63 + 13/99
A = 13/2 ( 2/15 + 2/35 + 2/63 + 2/99)
A= 13/2 ( 2/ 3.5 + 2/5.7 + 2/7.9 + 2/9.11)
A= 13/2 ( 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11)
A= 13/2 ( 1/3 - 1/11)
A= 13/2 . 8/33
A= 52/33
25 tháng 6 2018
\(\frac{31}{3}+\frac{31}{15}+\frac{31}{35}+\frac{31}{63}+\frac{31}{99}+\frac{31}{143}=\frac{31}{1.3}+\frac{31}{3.5}+\frac{31}{5.7}+\frac{31}{7.9}+\frac{31}{9.11}+\frac{31}{11.13}\\ \)
\(=\frac{31}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{31}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{31}{2}.\left(1-\frac{1}{13}\right)=\frac{31}{2}.\frac{12}{13}=\frac{31.6}{13}=\frac{186}{13}\)
\(\Rightarrow x-\frac{186}{13}=\frac{9}{13}\Leftrightarrow x=\frac{195}{13}=15\)
NV
25 tháng 6 2018
\(x-\left(\frac{31}{3}+\frac{31}{15}+\frac{31}{35}+\frac{31}{63}+\frac{31}{99}+\frac{31}{143}\right)=\)\(\frac{9}{13}\)(1)
Đặt \(A=\frac{31}{3}+\frac{31}{15}+\frac{31}{35}+\frac{31}{63}+\frac{31}{99}+\frac{31}{143}\)
\(A=31\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\right)\)
\(\Rightarrow2A=31\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)\)
\(2A=31\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(2A=31\left(2-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(2A=31\left(2-\frac{1}{13}\right)\)
\(2A=31.\frac{25}{13}\)
\(2A=\frac{775}{13}\)
\(\Rightarrow A=\frac{775}{13}:2\)
\(A=\frac{775}{26}\)
Thay vào (1) ta có:
\(x-\frac{775}{26}=\frac{9}{13}\)
\(\Leftrightarrow x=\frac{9}{13}+\frac{775}{26}\)
\(\Leftrightarrow x=\frac{61}{2}\)
Có: \(x=\dfrac{1}{9}+\dfrac{8}{116}=\dfrac{1}{9}+\dfrac{2}{29}=\dfrac{47}{261}\)
\(x-\left(\dfrac{2+2+2+2}{3+15+35+63}\right)=\dfrac{1}{9}\)
\(\Leftrightarrow x=\dfrac{1}{9}+\dfrac{2}{29}=\dfrac{47}{261}\)