Rút gọn:
a/ \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2000}\)
b/ \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{1998.1999.2000}\)
c/ \(C=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2006.2008}\)
Những câu hỏi liên quan
Afrac{1}{1.2}+frac{1}{2.3}+frac{1}{3.4}+...+frac{1}{2016.2017}Bfrac{1}{1.3}+frac{1}{3.5}+frac{1}{5.7}+...+frac{1}{2016.2017}Cfrac{1}{2.4}+frac{1}{4.6}+frac{1}{6.8}+...+frac{1}{2016.2018}Dfrac{1}{1.2.3}+frac{1}{2.3.4}+frac{1}{3.4.5}+...+frac{1}{98.99.100}Efrac{3}{4}.frac{8}{9}.frac{15}{16}cdot...cdotfrac{899}{900}F1.2+2.3+3.4+...+99.100MẤY BN NÀO BIẾT THÌ GIẢI JUP MK NHA!
Đọc tiếp
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
B=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2016.2017}\)
C=\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2016.2018}\)
D=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
E=\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}\cdot...\cdot\frac{899}{900}\)
F=1.2+2.3+3.4+...+99.100
MẤY BN NÀO BIẾT THÌ GIẢI JUP MK NHA!
Lâm đi là: 35 phút +2 giờ 20phút =2 giờ 55 phút
Đúng 0
Bình luận (0)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(A=1-\frac{1}{2017}\)
\(A=\frac{2016}{2017}\)
Đúng 0
Bình luận (0)
\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2016.2018}\)
\(2B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2016.2018}\)
\(2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2016}-\frac{1}{2018}\)
\(2B=1-\frac{1}{2018}\)
\(B=\frac{2017}{\frac{2018}{2}}\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
TÍNH TỔNG:
\(S=\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{2.3}-\frac{1}{2.3.4}+\frac{1}{3.4}-\frac{1}{3.4.5}+...+\frac{1}{99.100}-\frac{1}{99.100.101}\)
\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\right)\)
\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(=\left(1-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)
\(=\frac{99}{100}-\frac{1}{2}\cdot\frac{5049}{10100}=\frac{99}{100}-\frac{5049}{20200}=\frac{14949}{20200}\)
Đúng 0
Bình luận (0)
Tính tổng :
a/
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)=
b/
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...\frac{1}{28.29.30}\)=
c/
\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{27.28.29.30}\)=
a)\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+....+\(\frac{1}{100.101}\)=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+....+\(\frac{1}{100}\)-\(\frac{1}{101}\)=1-\(\frac{1}{101}\)=\(\frac{100}{101}\)
b)\(\frac{1}{1.2.3}\)+\(\frac{1}{2.3.4}\)+....+\(\frac{1}{28.29.30}\)=\(\frac{868}{3480}\)=\(\frac{217}{870}\)
c)\(\frac{1}{1.2.3.4}\)+\(\frac{1}{2.3.4.5}\)+....+\(\frac{1}{27.28.29.30}\)=\(\frac{24354}{438480}\)=\(\frac{451}{8120}\)
Đúng 0
Bình luận (0)
Tìm x biết
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2013.2014.2015}\right)x=\left(1.2+2.3+3.4+.....+2014.2015\right)\)
\(\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{2013.2014}-\frac{1}{2014.2015}\right)x=\frac{1}{3}\left(2014.2015.2016-2013.2014.2015........+2.3.4-1.2.3+1.2.3-0.1.2\right)\)
\(\left(\frac{1}{2}-\frac{1}{2014.2015}\right)x=\frac{1}{3}.2014.2015.2016\)
\(x=\frac{1}{3.2029104}.2014^2.2015^2.2016=\)
\(\left(\frac{1}{2}-\frac{1}{2014.2015}\right)x=\frac{1}{3}.2014.2015.2016\)
Đúng 0
Bình luận (0)
\(E=\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{2.3}-\frac{1}{2.3.4}+\frac{1}{3.4}-\frac{1}{3.4.5}+....+\frac{1}{99.100}-\frac{1}{99.100.101}\)
\(F=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{47.48.49.50}\)
Tính
\(C=1+\frac{1}{\left(-3\right)}+\frac{1}{\left(-3\right)^2}+....+\frac{1}{\left(-3\right)^{2015}}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{1998.1999.2000}\)
hãy tính tổng trên
Đặt tổng trên là A.Ta có:
A=\(\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{1998.1999.2000}\right)\)
A=\(\frac{1}{2}\left[\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\left(\frac{1}{1998.1999}-\frac{1}{1999.2000}\right)\right]\)
A=\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{1999.2000}\right)\)
A=\(\frac{1}{4}-\frac{1}{1999.2000.2}\)
Đúng 0
Bình luận (0)
\(\frac{1}{5678}\)
Ai ấn Đúng 0 sẽ may mắn cả năm đấy
Đúng 1
Bình luận (0)
\(F=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=\frac{n-1}{n}\)
\(G=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\left(n-1\right)\left(n-1\right).n}=\)
\(H=2+4+6+..+2n=\)
\(F=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=\frac{n-1}{n}\)
\(\Rightarrow F=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{\left(n-1\right)}-\frac{1}{n}\)
\(\Rightarrow F=1-\frac{1}{n}=\frac{n}{n}-\frac{1}{n}=\frac{n-1}{n}\left(đpcm\right)\)
\(H=2+4+6+...+2n\)
Đúng 0
Bình luận (0)
a) \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ \(\frac{1}{4.5}\)+ ..... +\(\frac{1}{99.100}\)
b) \(\frac{1}{1.2.3}\)+ \(\frac{1}{2.3.4}\)+ \(\frac{1}{3.4.5}\)+ .......+ \(\frac{1}{98.99.100}\)
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
b)\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{9900}=\frac{4949}{9900}\)
Đúng 0
Bình luận (0)
a)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
b)
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+....+\frac{1}{98\cdot99\cdot100}\)
\(=\frac{3-1}{1\cdot2\cdot3}+\frac{4-2}{2\cdot3\cdot4}+\frac{5-3}{3\cdot4\cdot4}+....+\frac{100-98}{98\cdot99\cdot100}\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+....+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\right)\)
\(=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)
Đúng 0
Bình luận (0)
a.
Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)
\(\Rightarrow A=\left[\frac{1}{1}-\frac{1}{2}\right]+\left[\frac{1}{2}-\frac{1}{3}\right]+\left[\frac{1}{3}-\frac{1}{4}\right]+...+\left[\frac{1}{99}-\frac{1}{100}\right]\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}\)
b.
\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\)
\(2B=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\)
\(2B=\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right]+\left[\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right]+...+\left[\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right]\)
\(2B=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\)
\(2B=\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\)
\(2B=\frac{1}{2}-\frac{1}{9900}=\frac{4949}{9900}\)
\(\Rightarrow B=\frac{4949}{9900}:2=\frac{4949}{19800}\)
Đúng 0
Bình luận (0)
Tính các tổng :a) Afrac{1}{1.2}+frac{1}{2.3}+frac{1}{3.4}+...+frac{1}{nleft(n+1right)} ( Hướng dẫn : frac{1}{kleft(k+1right)}frac{1}{k}-frac{1}{k+1})b) Bfrac{1}{1.2.3}+frac{1}{2.3.4}+frac{1}{3.4.5}+...+frac{1}{nleft(n+1right)left(n+2right)} ( Hướng dẫn : frac{1}{kleft(k+1right)left(k+2right)}frac{1}{2}left(frac{1}{k}+frac{1}{k+2}right)-frac{1}{k+1})
Đọc tiếp
Tính các tổng :
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\) ( Hướng dẫn : \(\frac{1}{k\left(k+1\right)}=\frac{1}{k}-\frac{1}{k+1}\))
b) \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
( Hướng dẫn : \(\frac{1}{k\left(k+1\right)\left(k+2\right)}=\frac{1}{2}\left(\frac{1}{k}+\frac{1}{k+2}\right)-\frac{1}{k+1}\))
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\)
\(A=1-\frac{1}{n+1}\)
a) Ta có: \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\)
\(A=1-\frac{1}{n+1}\)
\(A=\frac{n+1}{n+1}-\frac{1}{n+1}\)
\(A=\frac{n}{n+1}\)
Học tốt nha^^
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\)
\(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow2B=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow B=\frac{1}{4}-\frac{1}{2\left(n+1\right)\left(n+2\right)}\)