Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có :
\(S=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+..............+\dfrac{1}{98.99.100}\)
\(S=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+................+\dfrac{2}{98.99.100}\right)\)
\(S=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...........+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)
\(S=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{99.100}\right)\)
\(S=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{9900}\right)\)
\(S=\dfrac{1}{2}.\dfrac{4949}{9900}\)
\(S=\dfrac{4949}{19800}\)
~ Chúc bn học tốt ~
* Chứng tỏ
Ta có :\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\)
= \(\dfrac{1}{1.2.3}.\dfrac{2}{2}+\dfrac{1}{2.3.4}.\dfrac{2}{2}+...+\dfrac{1}{98.99.100}.\dfrac{2}{2}\)
= \(\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{98.99.100}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}+0+0+...+0+\dfrac{-1}{99.100}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{2}+\dfrac{-1}{9900}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{4850}{9900}+\dfrac{-1}{9900}\right)\)
= \(\dfrac{1}{2}.\dfrac{4849}{9900}\)
= \(\dfrac{4849}{19800}\)
a) Ta có: \(3xy+x-3y=6\)
\(\Rightarrow x\left(3y+1\right)-3y=6\)
\(\Rightarrow x\left(3y+1\right)-\left(3y+1\right)=5\)
\(\Rightarrow\left(x-1\right)\left(3y+1\right)=5\)
Ta có bảng sau:
....
b) Ta có: \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(=\frac{1}{2}.\frac{4949}{9900}\)
\(=\frac{4949}{19800}\)
\(\Rightarrow\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}=\frac{4949}{19800}\left(đpcm\right)\)
Vậy...
\(A=\left(\dfrac{-1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{4}{35}+\dfrac{2}{7}\right)+\dfrac{1}{127}\)
\(A=\left(\dfrac{-9-2-7}{18}\right)+\left(\dfrac{21+4+10}{35}\right)+\dfrac{1}{127}\)
\(A=-1+1+\dfrac{1}{127}\)
\(A=\dfrac{1}{127}\)
\(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\)
\(\dfrac{1}{4}B=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.4}+\dfrac{1}{3.4.5.4}+...+\dfrac{1}{98.99.100.4}\)
\(\dfrac{1}{4}B=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.\left(5-1\right)}+\dfrac{1}{3.4.5.\left(6-2\right)}+...+\dfrac{1}{98.99.100.\left(101-97\right)}\)
\(\dfrac{1}{4}B=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5-1.2.3.4}+\dfrac{1}{3.4.5.6-2.3.4.5}+...+\dfrac{1}{98.99.100.101-97.98.99.100}\)
\(\dfrac{1}{4}B=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}-\dfrac{1}{1.2.3.4}+\dfrac{1}{3.4.5.6}-\dfrac{1}{2.3.4.5}+...+\dfrac{1}{98.99.100.101}-\dfrac{1}{97.98.99.100}\)
\(\dfrac{1}{4}B=\dfrac{1}{98.99.100.101}\)
\(B=\dfrac{1}{98.99.100.101}.4=\dfrac{1}{98.99.25.101}\)
tick cho mk nha
bài tự làm 100%
co gì chưa đc thì coi lại nha
S = \(\dfrac{1}{1.4}\)+ \(\dfrac{1}{4.7}\)+...+\(\dfrac{1}{2002.2005}\)
S = ( 1 - \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-...+\(\dfrac{1}{2002}\)-\(\dfrac{1}{2005}\)) . \(\dfrac{1}{3}\)
S = ( 1 - \(\dfrac{1}{2005}\)) . \(\dfrac{1}{3}\)
S = \(\dfrac{2004}{2005}\). \(\dfrac{1}{3}\)
S = \(\dfrac{2014}{6015}\)
a) \(S=\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{2002.2005}\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2002}-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{3}.\dfrac{2004}{2005}=\dfrac{668}{2005}\)
KL.
b) \(P=\dfrac{3}{1.6}+\dfrac{3}{6.11}+\dfrac{3}{11.16}+...+\dfrac{3}{96.101}\)
\(=\dfrac{3}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{96}-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{5}\left(1-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{5}.\dfrac{100}{101}=\dfrac{60}{101}\)
KL.
c) \(Q=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{9900}\right)\)
\(=\dfrac{1}{2}.\dfrac{1}{19800}=\dfrac{1}{39600}\)
KL.
\(linh_1=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}\)
\(linh_1=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}\right)\)
\(linh_1=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{4.5}\right)\)
\(linh_1=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)=\dfrac{1}{2}.\dfrac{9}{20}=\dfrac{9}{40}\)
\(linh_2=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{8.9.10}\)
\(linh_2=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}\right)\)\(linh_2=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{9.10}\right)\)
\(linh_2=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{90}\right)=\dfrac{1}{2}.\dfrac{22}{45}=\dfrac{11}{45}\)
a/ \(G=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}\)
\(\Leftrightarrow2G=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}\)
\(\Leftrightarrow2G=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}\)
\(\Leftrightarrow2G=\dfrac{1}{1.2}-\dfrac{1}{4.5}\)
\(\Leftrightarrow2G=\dfrac{1}{2}-\dfrac{1}{20}\)
\(\Leftrightarrow2G=\dfrac{9}{20}\)
\(\Leftrightarrow G=\dfrac{9}{40}\)
b/ \(H=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+.....+\dfrac{1}{8.9.10}\)
\(\Leftrightarrow2H=\dfrac{2}{1.2.3}+\dfrac{2}{3.4.5}+.....+\dfrac{2}{8.9.10}\)
\(\Leftrightarrow2H=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+.....+\dfrac{1}{8.9}-\dfrac{1}{9.10}\)
\(\Leftrightarrow2H=\dfrac{1}{1.2}-\dfrac{1}{9.10}\)
\(\Leftrightarrow2H=\dfrac{1}{2}-\dfrac{1}{90}\)
\(\Leftrightarrow2H=\dfrac{22}{45}\)
\(\Leftrightarrow H=\dfrac{22}{90}\)
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{10.11.12}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{10.11.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{132}\right)\)
\(=\dfrac{1}{2}.\dfrac{65}{132}\)
\(=\dfrac{65}{264}\)
Vậy...
hôm qua cô giảng cho mình bài này không cần tính đâu
Gọi tổng là A
A=\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{17.18.19}\)
2A=\(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{17.18.19}\)
2A=\(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{17.18}-\dfrac{1}{18.19}\)
2A=\(\dfrac{1}{2}-\dfrac{1}{18.19}\)
A=\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{18.19}\right)\)
A=\(\dfrac{1}{2}.\dfrac{18.19-2}{2.18.19}\) < \(\dfrac{1}{4}\)
A=\(\dfrac{18.19-2}{2.2.18.19}\) < \(\dfrac{18.19}{2.2.18.19}\)
\(\Rightarrow\) A<\(\dfrac{1}{4}\)
\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{17.18.19}\)<\(\dfrac{1}{4}\)
Đặt A=\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{17.18.19}\)
2.A=2.(\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{17.18.19}\))
2. A=\(\dfrac{2}{1.2.3}\)+\(\dfrac{2}{2.3.4}\)+\(\dfrac{2}{3.4.5}\)+...+\(\dfrac{2}{17.18.19}\)
2.A=\(\dfrac{1}{1.2}\)-\(\dfrac{1}{2.3}\)+\(\dfrac{1}{2.3}\)-\(\dfrac{1}{3.4}\)+ ...+\(\dfrac{1}{17.18}\)-\(\dfrac{1}{18.19}\)
2.A=\(\dfrac{1}{1.2}\)-\(\dfrac{1}{18.19}\)=\(\dfrac{85}{171}\)
A=\(\dfrac{85}{171}\):2=\(\dfrac{85}{342}\)
Ta cũng có: \(\dfrac{1}{4}\) = \(\dfrac{171}{684}\); \(\dfrac{85}{342}\) = \(\dfrac{170}{684}\)
Vì 170 < 171 ( \(\dfrac{170}{684}\) < \(\dfrac{171}{684}\) )
Vậy \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{17.18.19}\) < \(\dfrac{1}{4}\)
E=\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\)
* Áp dụng công thức: \(\dfrac{k}{n.\left(n+k\right)}\)=\(\dfrac{1}{n}-\dfrac{1}{n+k}\)
ta có : \(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-....+\dfrac{1}{98.99}-\dfrac{1}{99.100}\)
E=\(\dfrac{1}{1.2}-\dfrac{1}{99.100}\)
E= ........(tính ra)
E=4949/9900