tìm x là số thỏa mãn
x+\(\dfrac{3}{4.7}\)+\(\dfrac{3}{7.10}\)+\(\dfrac{3}{10.13}\)+...+\(\dfrac{3}{37.40}\)=\(\dfrac{-37}{40}\)
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
VN
Võ Ngọc Phương
VIP
21 tháng 8 2023
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{34}{103}\)
\(\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{34}{103}\)
\(\dfrac{1}{3}.\left(1-\dfrac{1}{x+3}\right)=\dfrac{34}{103}\)
\(1-\dfrac{1}{x+3}=\dfrac{34}{103}:\dfrac{1}{3}=\dfrac{34}{103}.3\)
\(1-\dfrac{1}{x+3}=\dfrac{102}{103}\)
\(\dfrac{1}{x+3}=1-\dfrac{102}{103}=\dfrac{103}{103}-\dfrac{102}{103}\)
\(\dfrac{1}{x+3}=\dfrac{1}{103}\)
\(\Rightarrow x+3=103\)
\(x=103-3\)
\(x=100\)
Vậy x = 100
28 tháng 4 2018
1.
E = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{4.7}\) + \(\dfrac{3}{7.10}\) + \(\dfrac{3}{10.13}\) + \(\dfrac{3}{13.16}\) + \(\dfrac{3}{16.19}\) + \(\dfrac{3}{19.22}\)
E = 1 - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{10}\) + ... +\(\dfrac{1}{19}\) - \(\dfrac{1}{22}\)
E = 1 - \(\dfrac{1}{22}\)
E = \(\dfrac{21}{22}\)
2.
(x - 4)(x - 5) = 0
TH1:
x - 4 = 0 => x = 4
TH2:
x - 5 = 0 => x = 5
Vậy: x = 4 hoặc x = 5
27 tháng 4 2017
\(A=3.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\right)\)
\(A=3.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(A=3.\left(1-\dfrac{1}{100}\right)\)
\(A=3.\dfrac{99}{100}=\dfrac{297}{100}\)
28 tháng 8 2023
\(\dfrac{3}{1\times4}x+\dfrac{3}{4\times7}x+\dfrac{3}{7\times10}x+...+\dfrac{3}{31\times34}x=33\)
\(x\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+...+\dfrac{3}{31\times34}\right)=33\)
\(x\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)=33\)
\(x\left(1-\dfrac{1}{34}\right)=33\)
\(\dfrac{33}{34}x=33\)
\(x=34\)
VN
Võ Ngọc Phương
VIP
28 tháng 8 2023
\(\dfrac{3}{1.4}x+\dfrac{3}{4.7}x+\dfrac{3}{7.10}x+...+\dfrac{3}{31.34}x=33\)
\(x.3\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{31.34}\right)=33\)
\(x.3.\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)=33\)
\(x.\left(1-\dfrac{1}{34}\right)=33\)
\(x.\dfrac{33}{34}=33\)
\(x=33:\dfrac{33}{34}=33.\dfrac{34}{33}\)
\(x=34\)
TN
25 tháng 4 2018
\(\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{37.40}< \dfrac{1}{5}\)
=\(\dfrac{3}{3}\left(\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{37.40}\right)\)
=\(\dfrac{1}{3}\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{37.40}\right)\)
=\(\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{37}-\dfrac{1}{40}\right)\)
=\(\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{40}\right)\)
=\(\dfrac{3}{40}< \dfrac{1}{3}\)
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)
\(\dfrac{2}{42}+\dfrac{2}{56}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\\ \dfrac{2}{6.7}+\dfrac{2}{7.8}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\\ 2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\\ 2.\left(\dfrac{1}{6}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\\ \dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}=\dfrac{1}{18}\\ x+1=18\\ x=17.\)
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 1
\(\Leftrightarrow\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{375}{376}\)
\(\Leftrightarrow1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{375}{376}\)
\(\Leftrightarrow1-\dfrac{1}{x+3}=\dfrac{375}{376}\)
\(\Leftrightarrow\dfrac{1}{x+3}=1-\dfrac{375}{376}=\dfrac{1}{376}\)
\(\Rightarrow x+3=376\)
\(\Rightarrow x=373\)
1 tháng 6 2022
a: \(A=6\left(x+\dfrac{1}{3}\right)^2-7>=-7>-8\forall x\)
\(B=-8-\left(3.75-x\right)^2\le-8\)
Do đó: A>B
b: \(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}=\dfrac{15}{16}\)
\(B=\left(\dfrac{1}{2}\right)^4=\dfrac{1}{16}\)
Do đó: A>B
19 tháng 4 2017
Bài 1:
Ta có: \(A=\dfrac{2011+2012}{2012+2013}=\dfrac{2011}{2012+2013}+\dfrac{2012}{2012+2013}\)
Dễ thấy:
\(\dfrac{2011}{2012+2013}< \dfrac{2011}{2012};\dfrac{2012}{2012+2013}< \dfrac{2012}{2013}\)
\(\Rightarrow A=\dfrac{2011}{2012+2013}+\dfrac{2012}{2012+2013}< B=\dfrac{2011}{2012}+\dfrac{2012}{2013}\)
Bài 2:
\(S=\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{37\cdot40}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{37\cdot40}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{37}-\dfrac{1}{40}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{40}\right)=\dfrac{1}{3}\cdot\dfrac{9}{40}=\dfrac{3}{40}< \dfrac{1}{3}\)
\(x+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+...+\dfrac{3}{37\cdot40}=\dfrac{-37}{40}\\ x+\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+...+\dfrac{3}{37\cdot40}\right)=\dfrac{-37}{40}\\ x+\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{37}-\dfrac{1}{40}\right)=\dfrac{-37}{40}\\ x+\left(\dfrac{1}{4}-\dfrac{1}{40}\right)=\dfrac{-37}{40}\\ x+\dfrac{9}{40}=\dfrac{-37}{40}\\ x=\dfrac{-37}{40}-\dfrac{9}{40}\\ x=\dfrac{-46}{40}\\ x=\dfrac{-23}{20}\)
Vậy \(x=\dfrac{-23}{20}\)