Question

Giải giúp nhá mn!!!!!!

tìm kết quả của B: 

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

tick cho giải nhanh!!!!!!!!!!!!

 

 

# love is the beautiful pain #

Answer 1

#)Giải :

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

\(\Rightarrow5A=\frac{5}{4.9}+\frac{5}{9.14}+...+\frac{5}{44.49}\)

\(\Rightarrow5A=\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\)

\(\Rightarrow5A=\frac{1}{4}-\frac{1}{49}=\frac{45}{196}\)

\(\Rightarrow A=\frac{45}{196}\div5=\frac{9}{196}\)

Thay A vào B, ta được :

\(B=\frac{9}{196}.\frac{1-3-5-...-49}{89}\)

\(B=\frac{9}{196}.\frac{1-\left(3+5+7+...+49\right)}{89}\)

\(B=\frac{9}{196}.\frac{1-\left[\frac{\left(49+3\right).\left(\frac{49-3}{2}+1\right)}{2}\right]}{89}\)

\(B=\frac{9}{196}.\frac{-623}{89}=-\frac{9}{28}\)

Answer 2

B = \(\left(\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{44.49}\right).\frac{1-3-5-...-49}{89}\)

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

    = \(\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right).\frac{1-\left(24.52:2\right)}{89}\)

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

    = \(\frac{1}{5}.\frac{45}{196}.\left(-7\right)\)

    = \(\frac{-9}{28}\)

Vậy B = \(-\frac{9}{28}\)

Answer 3

=>5 B = (1/4-1/49) X  ( 1-(3+5+...+49))/89

=>5B= 45/196 X (-623)/89

=>5B=45/196 X (-7)

=> 5B = -45/28

=>B=-45/25:5

=>B=-9/28

   vẬY B=-9/28

Answer 4

  B = \(\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+...+\frac{1}{44.49}\right)\cdot\frac{1-3-...-49}{89}\)

=> 5B = \(5\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-3-...-49}{89}\)

=> 5B = \(\left(\frac{5}{4\cdot9}+\frac{5}{9\cdot14}+...+\frac{5}{44\cdot49}\right)\cdot\frac{1-3-...-49}{89}\)

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

=> 5B = \(\left(\frac{1}{4}-\frac{1}{49}\right)\cdot\frac{-623}{89}\)

=> 5B = \(\frac{45}{196}\cdot\left(-7\right)\)

=> B = \(\frac{-45}{28}:5\)= \(\frac{-9}{28}\)

# kiseki no enzeru #

hok tốt

Answer 5

trả lời:

\(B=\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.9}...+\frac{1}{44.49}\right).\frac{1-3-5-...-49}{89}\)

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

\(=\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right).\frac{1-\left(24.52:2\right)}{89}\)

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

\(=\frac{1}{5}.\frac{45}{196}.\left(-7\right)\)

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

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