K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

16 tháng 9 2017

\(\Rightarrow\left(\frac{1}{1}-\frac{1}{30}\right)x=-3\)

\(\Rightarrow\frac{29}{30}x=-3\)

\(\Rightarrow x=\left(-\frac{29}{90}\right)\)

16 tháng 9 2017

tính trog ngoặc trc nè :

\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{27.28.29.30}\)

=\(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)

=\(\frac{1}{1.2.3}-\frac{1}{28.29.30}\)

=\(\frac{1}{6}-\frac{1}{24360}\)

=\(\frac{1353}{8120}\)

thay vô biểu thức :

\(\frac{1353}{8120}.x=-3\)

x=\(-\frac{8120}{451}\)

16 tháng 1 2017

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}-3x=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{27.28.29.30}\)

\(\Leftrightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)-3x=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\right)\)

\(\Leftrightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)-3x=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right)\)

\(\Leftrightarrow\frac{4949}{19800}-3x=\frac{451}{8120}\)

\(\Leftrightarrow x\approx0,0648\)

14 tháng 1 2017

\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{27.28.29.30}\)

\(=\frac{1}{3}\left(\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...+\frac{3}{27.28.29.30}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right)\)

14 tháng 1 2017

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}-3x=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{27.28.29.30}\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)-3x=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\right)\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)-3x=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right)\)

\(\Rightarrow\frac{4949}{19800}-3x=\frac{451}{8120}\)

\(\Rightarrow3x=\frac{4949}{19800}-\frac{451}{8120}\)

\(\Rightarrow x=\left(\frac{4949}{19800}-\frac{451}{8120}\right):3\)

26 tháng 9 2021

Ta có \(\dfrac{1}{n\left(n+1\right)\left(n+2\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}=\dfrac{3}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)

Áp dụng:

\(\dfrac{1}{1\cdot2\cdot3\cdot4}+\dfrac{1}{2\cdot3\cdot4\cdot5}+...+\dfrac{1}{27\cdot28\cdot29\cdot30}\\ =\dfrac{1}{3}\left(\dfrac{3}{1\cdot2\cdot3\cdot4}+\dfrac{3}{2\cdot3\cdot4\cdot5}+...+\dfrac{3}{27\cdot28\cdot29\cdot30}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{1\cdot2\cdot3}-\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{2\cdot3\cdot4}-\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{27\cdot28\cdot29}-\dfrac{1}{28\cdot29\cdot30}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{1\cdot2\cdot3}-\dfrac{1}{28\cdot29\cdot30}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{6}-\dfrac{1}{24360}\right)=\dfrac{1}{3}\cdot\dfrac{1353}{8120}=\dfrac{451}{8120}\)

 

26 tháng 9 2021

\(\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{27.28.29.30}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+\dfrac{3}{3.4.5.6}+...+\dfrac{3}{27.28.29.30}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{27.28.29}-\dfrac{1}{28.29.30}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{1.2.3}-\dfrac{1}{28.29.30}\right)=\dfrac{1}{3}.\dfrac{4060-1}{28.29.30}\)

\(=\dfrac{1}{3}.\dfrac{4059}{24360}=\dfrac{1353}{24360}=\dfrac{451}{8120}\)

9 tháng 7 2017

Đặt \(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{17.18.19.20}\)

\(A=\frac{4-1}{1.2.3.4}+\frac{5-2}{2.3.4.5}+....+\frac{20-17}{17.18.19.20}\)

\(A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+....+\frac{3}{17.18.19.20}\)

\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+....+\frac{1}{17.18.19}-\frac{1}{18.19.20}\)

\(3A=\frac{1}{1.2.3}-\frac{1}{18.19.20}=\frac{1139}{6840}\)

\(\Rightarrow A=\frac{1139}{6840}\div3=\frac{1139}{20520}\)

8 tháng 10 2015

Lại phải giải hết 
Gọi dãy số trên là A
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+.....+\frac{1}{200.201.202.203}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-.....+\frac{1}{200.201.202}-\frac{1}{201.202.203}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{201.202.203}\)(chỗ này lm hơi tắt tí )
\(3A=\frac{1}{6}-\frac{1}{8242206}=\frac{1373701}{8242206}-\frac{1}{8242206}=\frac{1373700}{8242206}\)
\(A=\frac{1373700}{8242206}:3=\frac{457900}{8242206}\)

a) \(\frac{1}{2}-|\frac{5}{4}-2x|=\frac{1}{3}\Leftrightarrow|\frac{5}{4}-2x|=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{4}-2x=\frac{1}{6}\\\frac{5}{4}-2x=-\frac{1}{6}\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{5}{4}-\frac{1}{6}=\frac{13}{12}\\2x=\frac{5}{4}+\frac{1}{6}=\frac{17}{12}\end{cases}}}\)

Tự làm nốt và kết luận 

b) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)=0\)

Vì \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)\ne0\forall x\Rightarrow x+1=0\Leftrightarrow x=-1\)

Vậy ....