PH
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
16 tháng 7 2023
\(\dfrac{1}{15}\) + \(\dfrac{1}{21}\) + \(\dfrac{1}{28}\) + \(\dfrac{1}{36}\) +...+ \(\dfrac{2}{x\left(x+1\right)}\) = \(\dfrac{11}{40}\) (\(x\in\) N*)
\(\dfrac{1}{2}\).(\(\dfrac{1}{15}\)+\(\dfrac{1}{21}\)+\(\dfrac{1}{28}\)+\(\dfrac{1}{36}\)+.....+ \(\dfrac{2}{x\left(x+1\right)}\)) = \(\dfrac{11}{40}\) \(\times\) \(\dfrac{1}{2}\)
\(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\)+...+ \(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{11}{80}\)
\(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\)+...+ \(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{11}{80}\)
\(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)+...+ \(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\) = \(\dfrac{11}{80}\)
\(\dfrac{1}{5}\) - \(\dfrac{1}{x+1}\) = \(\dfrac{11}{80}\)
\(\dfrac{1}{x+1}\) = \(\dfrac{1}{5}\) - \(\dfrac{11}{80}\)
\(\dfrac{1}{x+1}\) = \(\dfrac{1}{16}\)
\(x\) + 1 = 16
\(x\) = 16 - 1
\(x\) = 15
NN
18 tháng 9 2020
\(\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+.......+\frac{2}{x\left(x+1\right)}=\frac{11}{40}\)
\(\Leftrightarrow\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+........+\frac{2}{x\left(x+1\right)}=\frac{11}{40}\)
\(\Leftrightarrow2.\left[\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+.......+\frac{1}{x\left(x+1\right)}\right]=\frac{11}{40}\)
\(\Leftrightarrow\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+......+\frac{1}{x\left(x+1\right)}=\frac{11}{80}\)
\(\Leftrightarrow\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+.......+\frac{1}{x\left(x+1\right)}=\frac{11}{80}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+.......+\frac{1}{x}-\frac{1}{x+1}=\frac{11}{80}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+1}=\frac{11}{80}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{16}\)
\(\Leftrightarrow x+1=16\)\(\Leftrightarrow x=15\)
Vậy \(x=15\)
NP
0
1 tháng 8 2016
b. (x+1)(1/10+1/11+1/12-1/13-1/14)=0
x+1=0 (vì : 1/10+1/11+1/12-1/13-1/14>0)
x=-1
5 tháng 8 2016
a) \(\frac{2}{\left(x+2\right).\left(x+4\right)}+\frac{4}{\left(x+4\right).\left(x+8\right)}+\frac{6}{\left(x+8\right).\left(x+14\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{x+14}{\left(x+2\right).\left(x+14\right)}-\frac{x+2}{\left(x+2\right).\left(x+14\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{x+14-x+2}{\left(x+2\right).\left(x+14\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{16}{\left(x+2\right).\left(x+4\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow x=16\)
Vậy x = 16
5 tháng 8 2016
\(b,\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\left(vì\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\right)\)
\(\Leftrightarrow x=-1\)
\(\text{Vậy }x=-1\)
12 tháng 8 2020
11/13-(5/42-x)=(15/28-11/13)
11/13-(5/42-x)=-37/182
(5/42-x)=11/13+37/182
(5/42-x)=191/182
x=5/42-191/182
x=-254/273
vậy x=-254/273
21 tháng 5 2015
\(\Rightarrow\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{x.\left(x+1\right)}=2.\left(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x.\left(x+1\right)}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}\right)=2.\left(\frac{1}{5}-\frac{1}{x+1}\right)=\frac{2}{5}-\frac{2}{x+1}=\frac{3}{10}\)
=> \(\frac{2}{x+1}\)= \(\frac{1}{10}=\frac{2}{20}\)
=> x +1 = 20 => x = 19
BK
24 tháng 7 2017
bạn trên sai rồi, nếu đã nhân đôi lên tất cả thì cx phải nhân luôn con cuối chứ
MH
2
9 tháng 8 2017
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{2}{9}.\frac{1}{2}\)
\(\Rightarrow\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}\)
\(\frac{1}{x+1}=\frac{1}{18}\Rightarrow18.1=1\left(x+1\right)\)
\(\Rightarrow18=x+1\Rightarrow x=18-1=17\)
27 tháng 6 2015
2) => \(-\frac{5}{42}-x=-\frac{18}{28}\) => \(-x=\frac{5}{42}-\frac{18}{28}=\frac{10}{84}-\frac{54}{84}=-\frac{44}{84}\)
=> \(x=\frac{44}{84}=\frac{11}{21}\)
3) => \(x=-\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{15}\right)=-\left(\frac{10}{60}+\frac{6}{60}-\frac{4}{60}\right)=-\frac{12}{60}=-\frac{1}{5}\)
4) => \(\frac{x}{5}=\frac{2}{10}-\frac{1}{5}-\frac{7}{50}=\frac{1}{5}-\frac{1}{5}-\frac{7}{50}=-\frac{7}{50}\)
=> \(x=5.\frac{-7}{50}=-\frac{7}{10}\)
Nhân cả 2 vế với 1/2
\(\Leftrightarrow\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{11}{80}\)
\(\Leftrightarrow\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+\dfrac{8-7}{7.8}+\dfrac{9-8}{8.9}+...+\dfrac{\left(x+1\right)-x}{x\left(x+1\right)}=\dfrac{11}{80}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{11}{80}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{x+1}=\dfrac{11}{80}\)
\(\Leftrightarrow16\left(x+1\right)-80=11\left(x+1\right)\Rightarrow x=15\)