Bạn chưa đăng nhập. Vui lòng đăng nhập để hỏi bài

Những câu hỏi liên quan
Phước Nguyễn
Xem chi tiết
Hồ Thu Giang
14 tháng 7 2015 lúc 8:45

\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

2\(\left(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)

\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2}{9}:2=\frac{1}{9}\)

\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}\)

=> x+1 =18

=> x = 18 - 1

=> x = 17

Nguyễn Phan Quỳnh Hương
Xem chi tiết
Trương Ứng Hòa
Xem chi tiết
Đỗ Thị Minh Anh
Xem chi tiết
long đức
Xem chi tiết
Hà Anh
Xem chi tiết
BAN is VBN
18 tháng 3 2016 lúc 20:17

\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow\frac{2}{6\cdot7}+\frac{2}{7\cdot8}+\frac{2}{8\cdot9}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\Rightarrow2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}\div2\)

\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{18}\)

\(\Rightarrow x+1=18\Rightarrow x=18-1\Rightarrow x=17\)

Phạm Thảo Vy
Xem chi tiết
soyeon_Tiểu bàng giải
27 tháng 9 2016 lúc 18:10

\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}\right)=\frac{1}{2}.\frac{2}{9}\)

\(\Rightarrow\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x.\left(x+1\right)}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x.\left(x+1\right)}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)

\(\Rightarrow\frac{1}{x+1}=\frac{3}{18}-\frac{2}{18}=\frac{1}{18}\)

\(\Rightarrow x+1=18\)

\(\Rightarrow x=18-1=17\)

Vậy \(x=17\)

tao lao
Xem chi tiết
Thanh Tùng DZ
3 tháng 6 2017 lúc 7:40

\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

\(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

\(2.\left(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2}{9}\)

\(2.\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x-1}\right)=\frac{2}{9}\)

\(2.\left(\frac{1}{6}-\frac{1}{x-1}\right)=\frac{2}{9}\)

\(\frac{1}{6}-\frac{1}{x-1}=\frac{2}{9}:2\)

\(\frac{1}{6}-\frac{1}{x-1}=\frac{1}{9}\)

\(\frac{1}{x-1}=\frac{1}{6}-\frac{1}{9}\)

\(\frac{1}{x-1}=\frac{1}{18}\)

\(\Rightarrow x-1=18\)

\(\Rightarrow x=18+1\)

\(\Rightarrow x=19\)

nguyển văn hải
3 tháng 6 2017 lúc 7:45

 = 2/42 + 2/56+2/72+................+2/x.(x+1)=2/9

=\(\frac{2}{6.7}\)+\(\frac{2}{7.8}\)+\(\frac{2}{8.9}\)+......+\(\frac{2}{x.\left(x+1\right)}\)=2/9

=2.( \(\frac{1}{6}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{8}\)+.......+\(\frac{1}{x}\)-\(\frac{1}{x+1}\)

=2.(1/6 -\(\frac{1}{x+1}\))=2/9

=1/6 -\(\frac{1}{x+1}\)=2/9:2=1/9

=1/6-1/9=\(\frac{1}{x+1}\)=3/54=1/18

=> x= 18-1 =17

nguyển văn hải
3 tháng 6 2017 lúc 8:10

BẠN SKT_NTT sai đề mà

Phan Đoàn Bảo Ngọc
Xem chi tiết
Dũng Nguyễn Đình
20 tháng 4 2016 lúc 17:48

Ta có : \(\frac{1}{21}+\frac{1}{28}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow2\left(\frac{1}{42}+\frac{1}{56}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)

\(\Rightarrow2\left(\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)

\(\Rightarrow2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\Rightarrow2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}:2=\frac{1}{9}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}\)

\(\Rightarrow x+1=18\Rightarrow x=17\)

Vậy x = 17

 

Ely Trần
9 tháng 4 2017 lúc 18:50

\(\Rightarrow\dfrac{2}{42}+\dfrac{2}{56}+\dfrac{2}{72}+.....+\dfrac{2}{x\left(x+1\right)}\Rightarrow2\left(\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+.....+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2}{9}\\ \Rightarrow2\left(\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+....+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2}{9}\\ \Rightarrow2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+....+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\\ \Rightarrow2\left(\dfrac{1}{6}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\\ \Rightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{2}{9}:2\\ \Rightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{2}{9}.\dfrac{1}{2}\\ \Rightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\\ \Rightarrow\dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\\ \Rightarrow\dfrac{1}{x+1}=\dfrac{3}{54}\\ \Rightarrow x+1=\dfrac{54}{3}\\ \Rightarrow x=\dfrac{54}{3}-1=\dfrac{51}{3}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \)

Phạm Lê Quỳnh Nga
20 tháng 4 2016 lúc 17:37

Làm tiếp :

​2. ( 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + ... + 1/x - 1/ x + 1) = 2/9

​2. ( 1/6 - 1/ x + 1) = 2/9

​1/6 - 1/x + 1 = 2/9 : 2

​1/6 - 1/ x + 1 = 1/9

1/ x + 1 = 1/6 - 1/9

1 : x + 1 = 1/18

x + 1 = 18

x = 18- 1 = 17