#)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 :
\(B=\frac{9}{196}.\frac{1-3-5-...-49}{89}\)
\(B=\frac{9}{196}.\frac{1-\left(3+5+7+...+49\right)}{89}\)
\(B=\frac{9}{196}.\frac{1-\left[\frac{\left(49+3\right).\left(\frac{49-3}{2}+1\right)}{2}\right]}{89}\)
\(B=\frac{9}{196}.\frac{-623}{89}=-\frac{9}{28}\)
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 = \(-\frac{9}{28}\)
=>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
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
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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