Cho A=\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{99\times100}\)
So sánh A với 1
\(\dfrac{2}{1\times2}+\dfrac{2}{2\times3}+\dfrac{2}{3\times4}+.....+\dfrac{2}{99\times100}\)
\(\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{99.100}\)
\(=2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)
\(=2.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=2.\left(1-\dfrac{1}{100}\right)=2.\dfrac{99}{100}=\dfrac{99}{50}\)
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+...+\dfrac{1}{98\times99}+\dfrac{1}{99\times100}\)
\(\dfrac{2}{11\times13}+\dfrac{2}{13\times15}+...+\dfrac{2}{19\times21}+\dfrac{2}{21\times23}\)
\(1,\\ =\dfrac{2-1}{1\times2}+\dfrac{3-2}{2\times3}+\dfrac{4-3}{3\times4}+\dfrac{5-4}{4\times5}+.....+\dfrac{99-98}{98\times99}+\dfrac{100-99}{99\times100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{100-1}{100}=\dfrac{99}{100}\)
\(2,=\dfrac{13-11}{11\times13}+\dfrac{15-13}{13\times15}+....+\dfrac{21-19}{19\times21}+\dfrac{23-21}{21\times23}\\ =\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+....+\dfrac{1}{19}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{23}\\ =\dfrac{1}{11}-\dfrac{1}{23}\\ =\dfrac{23-11}{11\times23}=\dfrac{12}{253}\)
@seven
a: 1/1*2+1/2*3+...+1/99*100
=1-1/2+1/2-1/3+...+1/99-1/100
=1-1/100
=99/100
b: 2/11*13+2/13*15+...+2/21*23
=1/11-1/13+1/13-1/15+...+1/21-1/23
=1/11-1/23
=12/253
\(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+...+\dfrac{2}{48\times49\times50}\)
\(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+...+\dfrac{2}{48\times49\times50}\)
\(=\dfrac{1}{1\times2}-\dfrac{1}{2\times3}+\dfrac{1}{2\times3}-\dfrac{1}{3\times4}+\dfrac{1}{3\times4}-\dfrac{1}{4\times5}+...+\dfrac{1}{48\times49}-\dfrac{1}{49\times50}\)
\(=\dfrac{1}{1\times2}-\dfrac{1}{49\times50}\)
\(=\dfrac{1}{2}-\dfrac{1}{2450}\)
\(=\dfrac{612}{1225}\)
\(\text{#}Toru\)
\(D=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+....+\dfrac{1}{2022\times2023}\)
D=1/1-1/2+1/2-1/3+1/3-1/4+.....+1/2022-1/2023
=1-1/2023=2022/2023
1/1x2+1/2x3+1/3x4+...+1/2022x2023
=1-1/2+1/2-1/3+1/3-1/4+....+1/2022-1/2023
=1-1/2023=2022/2023
...
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
1/* Chứng minh rằng:
\(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+...\dfrac{1}{49\times50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+..+\dfrac{1}{50}\)
2/* Cho:
A=\(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+.....+\dfrac{1}{99\times100}\). Chứng minh rằng:\(\dfrac{7}{12}< A>\dfrac{5}{6}\)
Các bn giúp mk những bài này nha!
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+\dfrac{x+3}{2001}-\dfrac{x+2}{2002}-\dfrac{x+1}{2003}=0\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1-\dfrac{x+2}{2002}-1-\dfrac{x+1}{2003}-1=0\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow x+2004\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\)
\(\Rightarrow x=-2004\)
Vậy \(x=-2004\)
1/ Ta có :
\(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+....+\dfrac{1}{49\times50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+....+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+.....+\dfrac{1}{50}\right)\)
\(\Rightarrow\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{50}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{50}\right)\times2\)
\(\Rightarrow\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{25}\right)\)
\(\Rightarrow\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+.....+\dfrac{1}{50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+.....+\dfrac{1}{50}\)
Hay \(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+...+\dfrac{1}{49\times50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\)
~ Học tốt nha ~
chứng minh
M=\(\dfrac{3}{1^2\times2^2}+\dfrac{5}{2^2\times3^2}+\dfrac{7}{3^2\times4^2}+.......+\dfrac{19}{9^2\times10^2}< 1\)
\(M=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(M=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)
\(M=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(M=1-\dfrac{1}{10^2}< 1\left(đpcm\right)\)
Tính :
a ) S= 2+4+6+...+2018 ( giải bằng hai cách )
b ) 10 + 102 +103 +...+10100 ( giaỉ bằng hai cách )
c ) \(S=\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{100}}\)( giải bằng hai cách )
d ) \(S=\dfrac{1!}{3!}+\dfrac{2!}{4!}+\dfrac{3!}{5!}+....+\dfrac{2018!}{2020!}\)
biết rằng : n! = \(1\times2\times3\times...\times n\)
VD : 1! = 1
2! = \(1\times2\)
3! = \(1\times2\times3\)
4! \(1\times2\times3\times4\)
a: Số số hạng là \(\dfrac{2018-2}{2}+1=1009\left(số\right)\)
Tổng là: \(\dfrac{2018+2}{2}\cdot1009=1009\cdot1010=1019090\)
b: \(10S=10^2+10^3+...+10^{101}\)
\(\Rightarrow9S=10^{101}-10\)
hay \(S=\dfrac{10^{101}-10}{9}\)
c: \(5S=1+\dfrac{1}{5}+...+\dfrac{1}{5^{99}}\)
\(\Leftrightarrow4S=1-\dfrac{1}{5^{100}}\)
hay \(S=\dfrac{1}{4}\left(1-\dfrac{1}{5^{100}}\right)\)
Tính bằng cách hợp lí :
a , \(\dfrac{1}{15}+\dfrac{9}{10}+\dfrac{14}{15}-\dfrac{11}{9}-\dfrac{20}{10}+\dfrac{1}{157}\)
b , \(\dfrac{1}{5}-\dfrac{-1}{3}+\dfrac{-1}{5}-\dfrac{2}{6}\)
c , \(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+...+\dfrac{2}{2015\times2017}\)
d , \(\dfrac{5}{1\times3}+\dfrac{5}{3\times5}+...+\dfrac{5}{2015\times2017}\)
e , \(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+...+\dfrac{1}{2016\times2017}\)
a: \(=\left(\dfrac{1}{15}+\dfrac{14}{15}\right)+\left(\dfrac{9}{10}-2-\dfrac{11}{9}\right)+\dfrac{1}{157}\)
\(=1+\dfrac{1}{157}+\dfrac{81-180-110}{90}\)
\(=\dfrac{158}{157}+\dfrac{-209}{90}\simeq-1.315\)
b: \(=\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{2}{6}\)
=1/3-1/3
=0
c: \(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2015\cdot2017}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)
=2016/2017
Bài 1: Tính
a) \(\dfrac{2\times3\times6\times7}{12\times7\times9\times2}\)
b) \(\dfrac{3\times4\times30\times56}{9\times8\times7\times8\times20}\)
(Các bạn tách ra rồi tính)