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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 \(\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+....+\frac{1}{44.49}\right).\frac{1-3-5-,,,,,-49}{89}=A.B=\frac{9}{196}.\left(-7\right)=-\frac{9}{28}\)
- #Ngân giải luôn nà. ^^ không gi lại đề nhá
\(=\left(\frac{5}{4\cdot9}+\frac{5}{9\cdot14}+\frac{5}{14\cdot19}+...+\frac{5}{44\cdot49}\right)\cdot\frac{1-\left(3+5+7+...+49\right)}{89}\)
\(=\frac{1}{5}\cdot\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right)\cdot\left(\frac{1-\frac{\left(52\cdot24\right)}{2}}{89}\right)\)
\(=\frac{1}{5}\cdot\left(\frac{1}{4}-\frac{1}{49}\right)\cdot\frac{1-624}{89}\)
\(=\frac{1}{5}\cdot\frac{45}{196}\cdot-\frac{623}{89}\)
\(=-\frac{9}{28}\)
#ĐÚng thì cho ngân ****+điểm
Đặt \(A=\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{44.49}\)
\(5A=\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\)
\(5A=\frac{1}{4}-\frac{1}{49}\)
\(A=\frac{45}{196}:5=\frac{9}{196}\)
Đặt \(B=\frac{1-3-...-49}{89}\)
\(B=\frac{\left(1-3\right)-\left(5-7\right)-...-\left(47-49\right)}{89}\)
\(B=\frac{\left(-2\right)-\left(-2\right)-...-\left(-2\right)}{89}\)
\(B=\frac{-2+2+...+2}{89}\)
\(B=\frac{\left(-2\right)+2\times24}{89}\)
\(B=\frac{46}{89}\)
\(P=A.B=\frac{9}{196}.\frac{46}{89}\)
\(P=\frac{207}{8722}\)
\(A=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+.....+\frac{1}{44}-\frac{1}{49}\right).\frac{1-\left(49+3\right)\left(\left(49-3\right):2+1\right):2}{89}\)
\(A=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right).\frac{1-26.24}{89}=\frac{45}{4.5.49}.\frac{-623}{89}=-\frac{9}{28}\)
\(A=\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+....+\frac{1}{44.49}\right).\frac{1-3-5-7-...-49}{89}\)
\(\Rightarrow5A=5.\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\right).\frac{1-3-5-7-...-49}{89}\)
\(=\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{44.49}\right).\frac{1+\frac{\left(-3-47\right).23}{2}-49}{89}\)
\(=\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right).\frac{1+\left(-575\right)-49}{89}\)
\(=\left(\frac{1}{4}-\frac{1}{49}\right).\frac{-623}{89}=\frac{45}{196}.\left(-7\right)=-\frac{45}{26}\)
\(\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+....+\frac{1}{44.49}\right)\cdot\frac{1-3-5-7-....-49}{89}\)
\(\text{Đặt }:\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\right)\)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:
\(\frac{\left(3+49\right).24}{2}=624\)
\(\Rightarrow\frac{1-624}{89}=\frac{-623}{89}=-7\)
\(\Rightarrow\left(\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{44.49}\right)\cdot\frac{1-3-5-7-...-49}{89}=A.B=\frac{9}{196}\cdot-7=-\frac{9}{28}\)
mk ko viết lại đề đâu
=\(\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right)\)\(.\frac{1-\left(3+5+...+49\right)}{89}\)
=\(\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right).\frac{\left(1-\frac{\left(49+3\right).24}{2}\right)}{89}\)
=\(\frac{1}{5}.\frac{45}{196}.\frac{1-\left(\frac{52.24}{2}\right)}{89}\)
=\(\frac{9}{196}.\left(1-\frac{624}{89}\right)=\frac{9}{196}.\left(\frac{-623}{89}\right)\)
=\(\frac{-9}{28}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right)\frac{2-\left(1+3+5+7+...+49\right)}{12}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\frac{2-\left(12.50+25\right)}{89}\)
\(=-\frac{5.9.7.89}{5.4.7.7.89}\)
\(=-\frac{9}{28}\)