tìm x,biết:
\(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+........+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
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Những câu hỏi liên quan
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 3 2022
\(\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{2}{42}+\dfrac{2}{56}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{2}{6.7}+\dfrac{2}{7.8}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\)
\(\Leftrightarrow x+1=18\)
\(\Leftrightarrow x=17\)
NA
Ngoc Anh Thai
Giáo viên
8 tháng 5 2021
a)
\(\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{9}{38}\\ \dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{9}{38}\\ \dfrac{1}{4}-\dfrac{1}{x+3}=\dfrac{9}{38}\\\\ \dfrac{1}{x+3}=\dfrac{1}{4}-\dfrac{9}{38}\\ \dfrac{1}{x+3}=\dfrac{1}{76}\\ x+3=76\\ x=73.\)
NA
Ngoc Anh Thai
Giáo viên
8 tháng 5 2021
b)
\(\dfrac{2}{42}+\dfrac{2}{56}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\\ \dfrac{2}{6.7}+\dfrac{2}{7.8}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\\ 2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\\ 2.\left(\dfrac{1}{6}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\\ \dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}=\dfrac{1}{18}\\ x+1=18\\ x=17.\)
6 tháng 5 2023
\(\dfrac{1}{3.7}\)+\(\dfrac{1}{7.4}\) +\(\dfrac{1}{4.9}\) +...+\(\dfrac{2}{x\left(x+1\right)}\) =\(\dfrac{2}{9}\)
\(\dfrac{2}{2.3.7}\)+\(\dfrac{2}{2.7.4}\) +\(\dfrac{2}{2.4.9}\) +...+\(\dfrac{2}{x\left(x+1\right)}\) =\(\dfrac{2}{9}\)
\(\dfrac{2}{6.7}\)+\(\dfrac{2}{7.8}\) +\(\dfrac{2}{8.9}\) +...+\(\dfrac{2}{x\left(x+1\right)}\) =\(\dfrac{2}{9}\)
2(\(\dfrac{1}{6.7}\) +\(\dfrac{1}{7.8}\) +\(\dfrac{1}{8.9}\) +...+\(\dfrac{1}{x\left(x+1\right)}\)) =\(\dfrac{2}{9}\)
2(\(\dfrac{1}{6}\) -\(\dfrac{1}{7}\) +\(\dfrac{1}{7}\) -\(\dfrac{1}{8}\) +\(\dfrac{1}{8}\) -\(\dfrac{1}{9}\) +...+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\) ) =\(\dfrac{2}{9}\)
2(\(\dfrac{1}{6}\) -\(\dfrac{1}{x+1}\) )=\(\dfrac{2}{9}\)
\(\dfrac{1}{6}\)-\(\dfrac{1}{x+1}\) =\(\dfrac{2}{9}\) : 2
\(\dfrac{1}{6}\)-\(\dfrac{1}{x+1}\) =\(\dfrac{1}{9}\)
\(\dfrac{1}{x+1}\) = \(\dfrac{1}{6}\) -\(\dfrac{1}{9}\)
\(\dfrac{1}{x+1}\) = \(\dfrac{1}{18}\)
x+1=18
x = 18-1
x =17
Vậy x =17
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
15 tháng 3 2017
Ta có:
\(A=\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow A=\dfrac{1}{3.7}+\dfrac{1}{4.7}+\dfrac{1}{4.9}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow A=\dfrac{1.2}{2.3.7}+\dfrac{1.2}{2.4.7}+\dfrac{1.2}{2.4.9}+...+\dfrac{1.2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow A=2\left(\dfrac{2}{6.7}+\dfrac{2}{7.8}+\dfrac{2}{8.9}+...+\dfrac{2}{x\left(x+1\right)}\right)=\dfrac{2}{9}\)
\(\Leftrightarrow A=2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\)
\(\Leftrightarrow A=2\left(\dfrac{1}{6}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\)
\(\Leftrightarrow x+1=18\)
\(\Leftrightarrow x=17\)
Vậy \(x=17\)
12 tháng 8 2017
\(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{1}{x\left(x+1\right):2}=\dfrac{2}{9}\)
<=> \(\dfrac{1}{6.7:2}+\dfrac{1}{7.8:2}+\dfrac{1}{8.9:2}+...+\dfrac{1}{x\left(x+1\right):2}=\dfrac{2}{9}\)
<=> \(\dfrac{2}{6.7}+\dfrac{2}{7.8}+\dfrac{2}{8.9}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
<=> \(2\left(\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2}{9}\)
<=> \(2\left(\dfrac{1}{6}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\)
<=> \(\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
<=> \(\dfrac{1}{x+1}=\dfrac{1}{18}\)
<=> x + 1 = 18
<=> x = 17
12 tháng 8 2017
\(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{1}{x.\left(x+1\right):2}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{1}{x\left(x+1\right):2}\right)=\dfrac{2}{9}.2=\dfrac{4}{9}\)\(\Leftrightarrow\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+...+\dfrac{1}{x}+\dfrac{1}{x+1}=\dfrac{4}{9}\)\(\Leftrightarrow\dfrac{1}{6}+\dfrac{1}{7}-\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{8}+...+\dfrac{1}{x}+\dfrac{1}{x+1}=\dfrac{4}{9}\)\(\Leftrightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{4}{9}-\dfrac{1}{6}=\dfrac{5}{8}\)
\(\Leftrightarrow\left(1.8\right)=5\left(x+1\right)\)
\(\Leftrightarrow8=5x+5\)
\(\Leftrightarrow5x=8-3=5\)
\(\Leftrightarrow x=5:5\)
\(\Leftrightarrow x=1\)
6 tháng 7 2017
a, sai đề
b, \(\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Rightarrow\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{9}\) ( nhân cả 2 vế với \(\dfrac{1}{2}\) )
\(\Rightarrow\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{9}\)
\(\Rightarrow\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\Rightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\Rightarrow x+1=18\Rightarrow x=17\)
Vậy x = 17
LG
6 tháng 7 2017
Câu a thiếu đề rồi bạn ơi mik giải câu b đây:
\(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\dfrac{2}{42}+\dfrac{2}{56}+\dfrac{2}{72}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\dfrac{2}{6.7}+\dfrac{2}{7.8}+\dfrac{2}{8.9}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+....+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\)
\(2\left(\dfrac{1}{6}-\dfrac{1}{x+2}\right)=\dfrac{2}{9}\)
\(\dfrac{1}{6}-\dfrac{1}{x+2}=\dfrac{2}{9}:2\)
\(\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\)
\(\dfrac{1}{x+1}=\dfrac{1}{18}\)
\(\Rightarrow x+1=18\Rightarrow x=17\)
Vậy x = 17
15 tháng 3 2017
Mấy bài này bạn tự làm đi, chuyển vế tìm x gần giống cấp I mà.
KW
20 tháng 3 2017
b)\(\dfrac{-3}{5}.x=\dfrac{1}{4}+0,75\)
=>\(\dfrac{-3}{5}.x=1\)
=>\(x=1:\dfrac{-3}{5}\)
=>\(x=\dfrac{-5}{3}\)
Vậy \(x=\dfrac{-5}{3}\)
\(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\\ \dfrac{2}{42}+\dfrac{2}{56}+\dfrac{2}{72}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\\ 2\cdot\left[\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+...+\dfrac{1}{x\left(x+1\right)}\right]=\dfrac{2}{9}\\ \dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2}{9}:2\\ \dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{9}\\ \dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{4}{9}\\ \dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\\ \dfrac{1}{x+1}=\dfrac{1}{18}\\ x+1=18\\ x=17\)
Vậy x = 17