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\(=\frac{1}{4}.\left(\frac{17.4}{1.3.5}+\frac{17.4}{3.5.7}+\frac{17.4}{5.7.9}+...+\frac{17.4}{47.49.51}\right)\)
\(=\frac{17}{4}\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{47.49}-\frac{1}{49.51}\right)\)
\(=\frac{17}{4}\left(\frac{1}{3}-\frac{1}{2499}\right)=\frac{17}{4}.\frac{832}{2499}=\frac{208}{147}\)
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(=>2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{98.99.100}\)
Dễ dàng CM đẳng thức phụ sau : \(\frac{2}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\)
Áp dụng vào tính 2B,ta có:
\(2B=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+....+\left(\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{1.2}-\frac{1}{99.100}=\frac{4949}{9900}=>B=\frac{4949}{9900}:2=\frac{4949}{19800}\)
Vậy.....
1/1.2.3 + 1/2.3.4 + .... + 1/98.99.100
= 1/2(1/1.2-1/2.3) + 1/2(1/2.3-1/3.4) + ..... + 1/2(1/98.99-1/99.100)
= 1/2(1/1.2-1/2.3+1/2.3-....+1/98.99-1/99.100)
= 1/2(1/2 - 1/9900)
= 1/2(4950/9900 - 1/9900)
= 1/2. 4949/9900
= 4949/19800
Bài làm:
Ta có: \(A=\frac{1}{1.3.5}+\frac{1}{3.5.7}+...+\frac{1}{47.49.51}\)
\(A=\frac{1}{4}\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{47.49.51}\right)\)
\(A=\frac{1}{4}\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{47.49}-\frac{1}{49.51}\right)\)
\(A=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{49.51}\right)\)
\(A=\frac{1}{12}-\frac{1}{4.49.51}< \frac{1}{12}\)
Vậy \(A< \frac{1}{12}\)
Từ đề bài suy ra\(4A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{47.49.51}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{47.49}-\frac{1}{49.51}=\frac{1}{3}-\frac{1}{49.51}< \frac{1}{3}\)
\(\Rightarrow A< \frac{1}{12}\left(đpcm\right)\)
a)|-10|:(-2):(-5)+(-3)2
=1+9
=10
b)1+(-2)+3+(-4)+5+(-6)+...+21+(-22)
=[1+(-2)]+[3+(-4)]+[5+(-6)]+...+[21+(-22]
=(-1)+(-1)+(-1)+...+(-1)
Mà từ 1 đến 22 có:(22-1):1+1:2=11(cặp)
Suy ra:1+(-2)+3+(-4)+5+(-6)+...+21+(-22)=(-11)
c)\(\frac{3}{4}.\frac{5}{9}+\frac{3}{4}.\frac{4}{9}\)
\(=\frac{3}{4}.\left(\frac{5}{9}+\frac{4}{9}\right)\)
\(=\frac{3}{4}\)
d)\(-\frac{4}{17}+\frac{5}{19}+-\frac{13}{17}+\frac{14}{19}+\frac{3}{115}\)
\(=\left[\left(-\frac{4}{17}\right)+\left(-\frac{13}{17}\right)\right]+\left(\frac{5}{19}+\frac{4}{19}\right)+\frac{3}{115}\)
\(=\left(-\frac{27}{17}\right)+1+\frac{3}{115}\)
\(=-\frac{1099}{1955}\)
e)\(\left(\frac{3}{4}+-\frac{7}{2}\right).\left(\frac{10}{11}+\frac{2}{22}\right)\)
\(=\left(\frac{3}{4}-\frac{14}{4}\right).\left(\frac{20}{22}+\frac{2}{22}\right)\)
\(=\left(-\frac{11}{4}\right).\left(\frac{22}{22}\right)\)
\(=-\frac{11}{4}\)
a) \(A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{98\cdot99\cdot100}\)
\(A=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\)
\(A=\frac{1}{2}-\frac{1}{99\cdot100}=\frac{1}{2}-\frac{1}{9900}=\frac{4949}{9900}\)
b) \(B=\frac{17}{1\cdot3\cdot5}+\frac{17}{3\cdot5\cdot7}+\frac{17}{5\cdot7\cdot9}+...+\frac{17}{47\cdot49\cdot51}\)
\(B=\frac{17}{4}\left(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+...+\frac{4}{47\cdot49\cdot51}\right)\)
\(B=\frac{17}{4}\left(\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{47\cdot49}-\frac{1}{49\cdot51}\right)\)
\(B=\frac{17}{4}\left(\frac{1}{3}-\frac{1}{2499}\right)=\frac{17}{4}\cdot\frac{832}{2499}=\frac{208}{147}\)