Tính tổng:
S1= 1/1.2 + 1/2.3 + 1/3.4 +...+ 1/47.48 + 1/48.49 + 1/49.50
S2= 1/4.7 + 1/7.10 + 1/10.13 +...+ 1/91.94 + 1/94.97 + 1/97.100
Giúp mình nha! Cảm ơn các bạn!😊
A=1/1.2+1/2.3+1/3.4+....+1/50.51
B=2/3.5+2/5.7+2/7.9+....+2/51.53
C=6/4.7+6/7.10+6/10.13+....+6/73.76
GIÚP MIK VỚI CẢM ƠN MN.
A = 1 /1.2 + 1/ 2.3 + 1 /3.4 + . . . + 1/ 49.50 + 1/ 50.51
A = 2 − 1/ 1.2 + 3 − 2 /2.3 + 4 − 3 /3.4 + . . . + 50 − 49 /49.50 + 51 − 50/ 50.51
A = 1 − 1/ 2 + 1/ 2 − 1 /3 + 1 /3 − 1/ 4 + . . . + 1 /50 − 1 /51
A=1-1/51
A=50/51
B=6/4.7+6/7.10+6/10.13+...+6/73.76
=2.(3/4.7 +3/7.10 +3/10.13 +...+3/73.76 )
=2.(1/4 −1/7 +1/7 −1/10 +1/10 −1/13 +...+1/73 −1/76 )
=2.(1/4-1/76)=2.9/38=9/19
Tính:
a/ -1/n - 1/n+a
b/ 1/1.2 + 1 /2.3 + 1/3.4 +......+ 1/2007.2008
c/ 3/1.4 + 3/4.7 + 3/7.10 +........+ 3/94.97
a; \(\dfrac{-1}{n}\) - \(\dfrac{1}{n+a}\)
= \(\dfrac{-n-a-n}{n.\left(n+a\right)}\)
= \(\dfrac{-2n-a}{n.\left(n+a\right)}\)
b; \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + ....+ \(\dfrac{1}{2007.2008}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2007}-\dfrac{1}{2008}\)
= \(\dfrac{1}{1}\) - \(\dfrac{1}{2008}\)
= \(\dfrac{2007}{2008}\)
c; \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)
= \(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)
= \(\dfrac{1}{1}\) - \(\dfrac{1}{97}\)
= \(\dfrac{96}{97}\)
Tính nhanh
a) 2/1.4+2/4.7+2/7.10+...+2/91.94+2/94.97
b) 1/5.8+1/8.11+1/11.14+...+1/605.608
(2/1.4=2 trên 1.4; 1/5.8=1 trên 5.8)
Ai lm đc nhớ nghi lời, cách giải đầy đủ
Cảm ơn!!!
\(\dfrac{3}{2}A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)
\(\dfrac{3}{2}A=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{97-94}{94.97}\)
\(\dfrac{3}{2}A=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{97}{94.97}-\dfrac{94}{94.97}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{97}=\dfrac{96}{97}\)
⇒ A = \(\dfrac{96}{97}:\dfrac{3}{2}=\dfrac{64}{97}\)
Câu B cách làm tương tự, thắc mắc gì bạn cứ hỏi nhé.
Tính:
a)\(\dfrac{1}{n}+\dfrac{1}{n+a}\)với a ;n là số tự nhiên và n khác 0
b) 1/1.2+1/2.3+1/3.4+...+
1/2008.2009
c)3/1.4+3/4.7+3/7.10+...+3/94.97
d)2/1.2+2/2.3+2/3.4+...+
2/2008.2009
Giải:
b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2008}-\dfrac{1}{2009}\)
\(=\dfrac{1}{1}-\dfrac{1}{2009}\)
\(=\dfrac{2008}{2009}\)
c) \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{4}{7.10}+...+\dfrac{3}{94.97}\)
\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)
\(=\dfrac{1}{1}-\dfrac{1}{97}\)
\(=\dfrac{96}{97}\)
Vậy ...
Các câu sau tương tự
b, \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{2008.1009}\)
\(=\)\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{2008}-\dfrac{1}{2009}\)
\(=\dfrac{1}{1}-\dfrac{1}{2009}=\dfrac{2009}{2009}-\dfrac{1}{2009}=\dfrac{2008}{2009}\)
c,\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+....+\dfrac{3}{94.97}\)
\(=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+....+\dfrac{97-94}{94.97}\)
\(=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{97}{94.97}-\dfrac{94}{94.97}\)
\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+....+\dfrac{1}{94}-\dfrac{1}{97}\)
\(=\dfrac{1}{1}-\dfrac{1}{97}=\dfrac{97}{97}-\dfrac{1}{97}=\dfrac{96}{97}\)
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2008.2009}\)
b) \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{94.97}\)
a)=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2009}\)
\(=1-\frac{1}{2009}\)
\(=\frac{2008}{2009}\)
b) =\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{94}-\frac{1}{97}\)
\(=1-\frac{1}{97}\)
=\(\frac{96}{97}\)
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2008.2009}\) \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2008}-\frac{1}{2009}\)
= 1 - 1/2009
= 2008/2009
b) 3/1.4 + 3/4.7 + 3/7.10 + .... + 3/94.97
= 1- 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + .... + 1/94 - 1/97
= 1 - 1/97
= 96/97
mình thống nhất là chịu
tk nhé@@@@@@@@@@@@@@@@@@@
hay không
nếu ban thích
so sánh 3/1.4+3/4.7+3/7.10+....+3/91.94+3/94.97 với 1
\(\frac{3}{1.4}+\frac{3}{4.7}+.....+\frac{3}{94.97}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.........+\frac{1}{94}-\frac{1}{97}\)
\(=1-\frac{1}{97}\)
\(=\frac{96}{97}\)
mà \(\frac{96}{97}< 1\)
\(\Rightarrow\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{94.07}< 1\)
vậy..................
\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{91\cdot94}+\frac{3}{94\cdot97}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}\)
\(=1-\frac{1}{97}\)
\(=\frac{96}{97}\)
\(\Rightarrow\frac{96}{97}< 1\)
\(\Rightarrow\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{94\cdot97}< 1\)
Vậy \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{94\cdot97}< 1\)
So sánh 3/1.4 + 3/4.7+ 3/7.10+...+3/91.94+3/94.97 với 1
Ta có: \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}\)
\(\Leftrightarrow1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{94}-\frac{1}{97}\)
\(\Leftrightarrow1-\frac{1}{97}=\frac{96}{97}\)
Do \(\frac{96}{97}< 1\Rightarrow\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{94.97}< 1\)
Vậy:.............................<1
Câu 1 : 1.2+2.3+3.4+...+30.31
Câu 2 : 1.2.3+2.3.4+3.4.5+...+30.31.32
Câu 3 : 1/1.2+1/2.3+1/3.4+...+1/30.31
Câu 4 ; 1/1.3+1/3.5+...+1/99.101
Câu 5 : 1/1.4+1/4.7+...+1/91.94
Câu 6 : 1/1.2.3+1/2.3.4+1/3.4.5+...+1/31.32.33
Câu 7 : 1.1!+2.2!+3.3!+...+10.10!
Tính:
a) \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) +...+ \(\dfrac{1}{1999.2000}\)
b) \(\dfrac{1}{1.4}\) + \(\dfrac{1}{4.7}\) + \(\dfrac{1}{7.10}\) +...+ \(\dfrac{1}{100+103}\)
c) \(\dfrac{8}{9}\) - \(\dfrac{1}{72}\) - \(\dfrac{1}{56}\) - \(\dfrac{1}{42}\) -...-\(\dfrac{1}{6}\) - \(\dfrac{1}{2}\)
`#3107`
`a)`
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{1999\cdot2000}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{1999}-\dfrac{1}{2000}\)
\(=1-\dfrac{1}{2000}\)
\(=\dfrac{1999}{2000}\)
`b)`
\(\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{100\cdot103}?\)
\(=\dfrac{1}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{100\cdot103}\right)\)
\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{102}{103}\)
\(=\dfrac{34}{103}\)
`c)`
\(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-....-\dfrac{1}{6}-\dfrac{1}{2}\)
\(=\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)\)
\(=\dfrac{8}{9}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)\)
\(=\dfrac{8}{9}-\dfrac{8}{9}\\ =0\)
b) Sửa đề:
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}\)
\(=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}.\left(1-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}.\left(\dfrac{103}{103}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}.\dfrac{102}{103}\)
\(=\dfrac{34}{103}\)
c) \(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-...-\dfrac{1}{6}-\dfrac{1}{2}\)
\(=\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)\)
\(=\dfrac{8}{9}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)\)
\(=\dfrac{8}{9}-\left(\dfrac{9}{9}-\dfrac{1}{9}\right)\)
\(=\dfrac{8}{9}-\dfrac{8}{9}\)
\(=0\)
\(#WendyDang\)