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25 tháng 12 2016

Hỏi thật không

10 tháng 2 2020

\(A=\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\frac{1-3-5-7-...-49}{89}\\ A=\frac{1}{5}\left(\frac{5}{4\cdot9}+\frac{5}{9\cdot14}+\frac{5}{14\cdot19}+...+\frac{5}{44\cdot49}\right)\frac{1-\left(3+5+7+...+49\right)}{89}\\ A=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right)\frac{1-\frac{\left(49+3\right)\cdot24}{2}}{89}\\ A=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\cdot\left(-7\right)\\ A=\frac{1}{5}\cdot\frac{45}{196}\cdot\left(-7\right)\\ A=\frac{-9}{28}\)

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

13 tháng 5 2016

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

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

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

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

Đặt \(B=\frac{1-3-5-7-.......47-49}{89}\)

\(B=\frac{1-\left(3+5+7+......+47+49\right)}{89}\)

Từ 3 -> 49 có: (49-3):2+1=24(số hạng)

=>\(3+5+7+....+47+49=\frac{\left(49+3\right).24}{2}=624\)

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

Vậy \(\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+....+\frac{1}{44.49}\right).\frac{1-3-5-,,,,,-49}{89}=A.B=\frac{9}{196}.\left(-7\right)=-\frac{9}{28}\)

12 tháng 5 2016

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right)\frac{2-\left(1+3+5+7+...+49\right)}{12}\)

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\frac{2-\left(12.50+25\right)}{89}\)

\(=-\frac{5.9.7.89}{5.4.7.7.89}\)

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

20 tháng 11 2018

\(=\left[\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}\right)+\frac{1}{5}\left(\frac{1}{9}-\frac{1}{14}\right)+\frac{1}{5}\left(\frac{1}{14}-\frac{1}{19}\right)+...+\frac{1}{5}\left(\frac{1}{44}-\frac{1}{49}\right)\right]\cdot\frac{1-\left(3+5+...+49\right)}{89}\)

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-...+\frac{1}{44}-\frac{1}{49}\right)\cdot\frac{1-\left(52+52+...+52\right)\left\{12\text{ số 52}\right\}}{89}\)

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\cdot\frac{1-624}{89}\)

\(=\frac{9}{196}\cdot-7=\frac{9}{28}\)