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28 tháng 4

\(x+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+...+\dfrac{3}{37\cdot40}=\dfrac{-37}{40}\\ x+\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+...+\dfrac{3}{37\cdot40}\right)=\dfrac{-37}{40}\\ x+\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{37}-\dfrac{1}{40}\right)=\dfrac{-37}{40}\\ x+\left(\dfrac{1}{4}-\dfrac{1}{40}\right)=\dfrac{-37}{40}\\ x+\dfrac{9}{40}=\dfrac{-37}{40}\\ x=\dfrac{-37}{40}-\dfrac{9}{40}\\ x=\dfrac{-46}{40}\\ x=\dfrac{-23}{20}\)

Vậy \(x=\dfrac{-23}{20}\)

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

28 tháng 4 2018

Cho mình hỏi là số ở đâu ra luôn đc ko bạn?

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

\(\dfrac{3}{1\times4}x+\dfrac{3}{4\times7}x+\dfrac{3}{7\times10}x+...+\dfrac{3}{31\times34}x=33\)

\(x\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+...+\dfrac{3}{31\times34}\right)=33\)

\(x\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)=33\)

\(x\left(1-\dfrac{1}{34}\right)=33\)

\(\dfrac{33}{34}x=33\)

\(x=34\)

28 tháng 8 2023

\(\dfrac{3}{1.4}x+\dfrac{3}{4.7}x+\dfrac{3}{7.10}x+...+\dfrac{3}{31.34}x=33\)

\(x.3\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{31.34}\right)=33\)

\(x.3.\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)=33\)

\(x.\left(1-\dfrac{1}{34}\right)=33\)

\(x.\dfrac{33}{34}=33\)

\(x=33:\dfrac{33}{34}=33.\dfrac{34}{33}\)

\(x=34\)

 

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
4 tháng 1

\(\Leftrightarrow\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{375}{376}\)

\(\Leftrightarrow1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{375}{376}\)

\(\Leftrightarrow1-\dfrac{1}{x+3}=\dfrac{375}{376}\)

\(\Leftrightarrow\dfrac{1}{x+3}=1-\dfrac{375}{376}=\dfrac{1}{376}\)

\(\Rightarrow x+3=376\)

\(\Rightarrow x=373\)

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:

\(S=\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{37\cdot40}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{37\cdot40}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{37}-\dfrac{1}{40}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{40}\right)=\dfrac{1}{3}\cdot\dfrac{9}{40}=\dfrac{3}{40}< \dfrac{1}{3}\)