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11 tháng 7 2017

\(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right)\)

\(=\) \(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\)

\(=\) \(\dfrac{1}{4}-\dfrac{1}{49}\)

\(=\) \(\dfrac{49}{196}-\dfrac{4}{196}\)

\(=\) \(\dfrac{45}{196}\)

11 tháng 7 2017

Biểu thức ban đầu không thỏa công thức nên không giải như vậy đc => sai.

30 tháng 1 2022

Đặt \(A=\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}\)

\(=\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+\dfrac{5}{14.19}+...+\dfrac{5}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}\)

\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right).\dfrac{1-3-5-7-...-49}{89}\)

\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right).\dfrac{1-3-5-7-...-49}{89}\)

\(=\dfrac{9}{196}.\dfrac{1-3-5-7-...-49}{89}\)

Đặt \(B=1-3-5-7-..-49\)

\(=1-\left(3+5+7+...+49\right)\)

\(=1-\left\{\left(49+3\right).\left[\left(49-3\right):2+1\right]:2\right\}\)

\(=1-624\)

\(=-623\)

\(\Rightarrow\dfrac{9}{196}.\left(\dfrac{-623}{89}\right)=-\dfrac{9}{28}\)

Vậy: \(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}=-\dfrac{9}{28}\)

30 tháng 1 2022

Xét \(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right)\)

=\(\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+\dfrac{5}{14.19}+...+\dfrac{5}{44.49}\right)\)

=\(\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)\)

=\(\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\)

=\(\dfrac{1}{5}.\dfrac{45}{196}\)

=\(\dfrac{9}{196}\)

Xét \(\dfrac{1-3-5-7-..-49}{89}\)

=\(\dfrac{1-\left(3+5+7+...+49\right)}{89}\)

CT tính sl số hạng (số cuối - số đầu ):2+1

số lượng số hạn của dãy 3+5+7+...+49 là (49-3):2+1=24

Áp dụng CT tính tổng số hạng dãy số cách đều Tổng = [ (số đầu + số cuối) x Số lượng số hạng ] : 2

=> tổng = [(3+49).24]:2=624

=>\(\dfrac{1-624}{89}\)

=\(\dfrac{-623}{89}\)

=-7

từ đó ta có \(\dfrac{9}{196}.\left(-7\right)=\dfrac{-9}{28}\)

Ta có: \(A=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{1}{5}\cdot\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+\dfrac{5}{14\cdot19}+...+\dfrac{5}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{1}{5}\cdot\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{1}{5}\cdot\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{1}{5}\cdot\left(\dfrac{49-4}{4\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{1}{5}\cdot\dfrac{45}{196}\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{9}{196}\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{9}{196}\cdot\dfrac{-623}{89}=-\dfrac{9}{28}\)

22 tháng 8 2023

\(A=\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+..+\dfrac{1}{44.49}\right)\left(\dfrac{1-3-5-7-..-49}{89}\right)\\ A=\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+..+\dfrac{5}{44.49}\right)\left(\dfrac{1-3-5-7-...-49}{89}\right)\\ A=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\left(\dfrac{1-3-5-7-...-49}{89}\right)\)

\(A=\dfrac{9}{196}\left(\dfrac{1-3-5-7-...-49}{89}\right)\)

Ta đặt: \(P=1-3-5-7-...-49\\ =1-\left(3+5+7+..+49\right)\\ =1-624\\ =-623\\ \Rightarrow\dfrac{9}{196}.-\dfrac{623}{89}=-\dfrac{9}{28}.\)

22 tháng 8 2023

Ta có: �=(14⋅9+19⋅14+114⋅19+...+144⋅49)⋅1−3−5−7−...−4989

⇔�=15⋅(54⋅9+59⋅14+514⋅19+...+544⋅49)⋅1−3−5−7−...−4989

⇔�=15⋅(14−19+19−114+114−119+...+144−149)⋅1−3−5−7−...−4989

⇔�=15⋅(14−149)⋅1−3−5−7−...−4989

⇔�=15⋅(49−44⋅49)⋅1−3−5−7−...−4989

⇔�=15⋅45196⋅1−3−5−7−...−4989

⇔�=9196⋅1−3−5−7−...−4989

⇔�=9196⋅−62389=−928
 

1 tháng 2 2017

\(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\)

\(=\frac{1}{5}\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{44.49}\right)\)

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right)\)

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)=\frac{9}{196}\)

1 tháng 2 2017

thực hiện phép tính:1-3-5-...-49/89

23 tháng 7 2017

\(\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+....+\frac{1}{44.49}\right)\cdot\frac{1-3-5-7-....-49}{89}\)

  \(\text{Đặt }:\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\right)\)là \(A\)

            \(\frac{1-3-5-7-...-49}{89}\)là \(B\);ta có : 

\(A=\frac{1}{5}\cdot\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right)\)

\(A=\frac{1}{5}\cdot\left(\frac{1}{4}-\frac{1}{49}\right)=\frac{1}{5}\cdot\frac{45}{196}=\frac{9}{196}\)

\(B=\frac{1-3-5-7-....-49}{89}=\frac{1-\left(3+5+7+...+49\right)}{89}\)

Tổng của \(3+5+7+...+49\)là: 

\(\frac{\left(3+49\right).24}{2}=624\)

\(\Rightarrow\frac{1-624}{89}=\frac{-623}{89}=-7\)

\(\Rightarrow\left(\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{44.49}\right)\cdot\frac{1-3-5-7-...-49}{89}=A.B=\frac{9}{196}\cdot-7=-\frac{9}{28}\)

25 tháng 3 2018

mk ko viết lại đề đâu

=\(\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right)\)\(.\frac{1-\left(3+5+...+49\right)}{89}\)

=\(\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right).\frac{\left(1-\frac{\left(49+3\right).24}{2}\right)}{89}\)

=\(\frac{1}{5}.\frac{45}{196}.\frac{1-\left(\frac{52.24}{2}\right)}{89}\)

=\(\frac{9}{196}.\left(1-\frac{624}{89}\right)=\frac{9}{196}.\left(\frac{-623}{89}\right)\)

=\(\frac{-9}{28}\)

30 tháng 12 2017

ta có

1/5(5/36+5/126+...+5/44*49)1-3-5-7-9-...-49/89

=1/5(1/4-1/9+1/9-1/14+...+1/44-1/49)-623/89

=1/5*-7(1/4-1/49)

=-7/5*45/196

=-9/128

31 tháng 12 2017

bạn ơi 9/28 chứ không phải 9/128 đâu

28 tháng 3 2017

=\(\dfrac{1}{5}\).(\(\dfrac{5}{4.9}+\dfrac{5}{9.14}+\dfrac{5}{14.19}+....+\dfrac{5}{44.49}\)).\(\dfrac{1-\left(3+5+7+...+49\right)}{89}\)

=\(\dfrac{1}{5}.\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)\).\(\dfrac{1-624}{89}\)

=\(\dfrac{1}{5}.\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\).(-7)

=\(\dfrac{1}{5}\).\(\dfrac{45}{196}\).(-7)=\(\dfrac{-9}{28}\)

29 tháng 3 2017

fty

b: Ta có: \(B=\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+...+\dfrac{1}{64\cdot69}\)

\(=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{64\cdot69}\right)\)

\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{64}-\dfrac{1}{69}\right)\)

\(=\dfrac{1}{5}\cdot\dfrac{65}{4\cdot69}\)

\(=\dfrac{13}{276}\)

26 tháng 9 2021

\(A=\dfrac{2}{1\cdot4}+\dfrac{2}{4\cdot7}+...+\dfrac{2}{97\cdot100}\\ A=\dfrac{2}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{97\cdot100}\right)\\ A=\dfrac{2}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\\ A=\dfrac{2}{3}\left(1-\dfrac{1}{100}\right)=\dfrac{2}{3}\cdot\dfrac{99}{100}=\dfrac{33}{50}\\ B=\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+...+\dfrac{1}{64\cdot69}\\ B=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{64\cdot69}\right)\\ B=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{64}-\dfrac{1}{69}\right)\\ B=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{69}\right)=\dfrac{1}{5}\cdot\dfrac{65}{276}=\dfrac{13}{276}\)

\(C=70\left(\dfrac{13}{56}+\dfrac{13}{72}+\dfrac{13}{90}\right)=70\cdot13\left(\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\\ C=910\left(\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\\ C=910\left(\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ C=910\left(\dfrac{1}{7}-\dfrac{1}{10}\right)=910\cdot\dfrac{3}{70}=39\)