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\(B=\frac{2.5-1}{2.5}+\frac{5.8-1}{5.8}+\frac{8.11-1}{8.11}+...+\frac{50.53-1}{50.53}\)
\(B=1-\frac{1}{2.5}+1-\frac{1}{5.8}+1-\frac{1}{8.11}+...+1-\frac{1}{50.53}\)
\(B=17-\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{50.53}\right)\)
\(B=17-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{50}-\frac{1}{53}\right)\)
\(B=17-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{53}\right)=\frac{1785}{106}\)
\(A=\frac{5}{2.5}+\frac{5}{5.8}+\frac{5}{8.11}+...+\frac{5}{98.101}\)
\(=\frac{5}{2}-\frac{5}{5}+\frac{5}{5}-\frac{5}{8}+....+\frac{5}{98}-\frac{5}{101}\)
\(=\frac{5}{2}-\frac{5}{101}=\frac{495}{202}\)
\(\frac{5}{2\times5}+\frac{5}{5\times8}+\frac{5}{8\times11}+...+\frac{5}{98\times101}\)
\(=\frac{5}{3}\times\left(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{98\times101}\right)\)
\(=\frac{5}{3}\times\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\right)\)
\(=\frac{5}{3}\times\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=\frac{5}{3}\times\frac{99}{202}=\frac{165}{202}\)
=> 3B = 3.( 1/2.5 + 1/5.8 + 1/8.11 + ........... + 1/122.125)
= 3/2.5 + 3/5.8 + 3/ 8.11 + ......+ 3/122.125
Ta có: 3/ 2.5 = 1/2 - 1/5
3/5.8 = 1/5 -1/8
3/ 8.11 = 1/8 -1/11
..........................
3/122 . 125 = 3/122 - 3/125
=> 3B= 1/2 - 15/5 + 1/5 -1/8 +1/8 - 1/11 +........+1/122 - 1/125
= 1/2 - 1/125 = 125/250 - 2/250= 123/250
=> B= 3B : 3 = 123/250 :3 = 123/250 . 1/3 = 41/250
=> 2C = 2.(1/9.11 + 1/11.13 +....+ 1/97 .99)
= 2/9.11 + 2/11 .13 +.....+ 2/ 97.99
Ta có: 2/9.11 = 1/9 - 1/11
2/11.13 = 2/11 -2/ 13
...............................
2/97.99 = 1/97 - 1/99
=> 2B = 1/9 - 1/11 + 1/11 - 1/13 + ....+ 1/97 - 1/99
= 1/9 -1/99 = 11/99 - 1/99 =10/99
=> B= 2B : B = 10/99 :2 =10/99 . 1/2 = 5/99
Vậy B = 5/99
\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{62.65}\)
\(=1.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{62}-\frac{1}{65}\right)\)
\(=1.\left(\frac{1}{2}-\frac{1}{65}\right)\)
\(=1.\frac{63}{130}\)
\(=\frac{63}{130}\)
Bài làm
\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{62.65}\)
\(=3.\frac{1}{2.5}+3.\frac{1}{5.8}+3.\frac{1}{8.11}+...+3.\frac{1}{62.65}\)
\(=3.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{62.65}\right)\)
\(=3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{62}-\frac{1}{65}\right)\)
\(=3.\left(\frac{1}{2}-\frac{1}{65}\right)\)
\(=3.\left(\frac{65}{130}-\frac{2}{130}\right)\)
\(=3.\frac{63}{130}\)
\(=\frac{3.63}{130}\)
\(=\frac{189}{130}\)
# Chúc bạn học tốt #
\(\frac{3}{2.5}\)+ \(\frac{3}{5.8}\)+ ...... + \(\frac{3}{92.95}\)= 3 . ( \(\frac{1}{2.5}\)+ \(\frac{1}{5.8}\)+ .... + \(\frac{1}{92.95}\))
= 3 . \(\frac{1}{3}\). ( \(\frac{1}{2.5}\)+ \(\frac{1}{5.8}\)+ ..... + \(\frac{1}{92.95}\))
= 3. \(\frac{1}{3}\). ( \(\frac{1}{2}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{8}\)+ ....... + \(\frac{1}{92}\)- \(\frac{1}{95}\))
= 1 .( \(\frac{1}{2}\)- \(\frac{1}{95}\)) = \(\frac{93}{190}\)
Thấy hay thì cho mình một k nhé!!!
3/ 2.5 + 3/ 5.8 + 3/ 8.11+ ...+ 3/ 92.95
=1/2-1/5+1/5-1/8+1/8-1/11+........+1/92-1/95
=1/2-1/95
=31/60
= 1/3.(1/2-1/5)+1/3.(1/5-1/8)+....+1/3.(1/92-1/95)+1/3.(1/95-1/98)
=1/3.(1/2-1/5+1/5-1/8+....+1/92-1/95+1/95-1/98)
=1/3.(1/2-1/98)
=1/3.24/49
=8/49
Phân tích: 1/2.5 = 1/2 - 1/5
1/5.8 = 1/5 - 1/8
1/8.11 = 1/8 - 1/11
...
1/92.95 = 1/92 - 1/95
1/95.98 = 1/95 - 1/98
Ta có: 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 +...+ 1/92 - 1/95 + 1/95 - 1/98
3 = 3/2.5 + 3/5.8 + 3/8.11 + ...+ 3/92.95 + 3/95.98
3 = 1 - 1/2 + 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 +...+ 1/92 - 1/95 + 1/95 - 1/98
= 1 - 1/98
= 97/98 : 3 = 97/98 x 1/3 = (tự tính)
Kham khảo tại link:
Câu hỏi của Lê Thúy Hằng - Toán lớp 6 | Học trực tuyến
https://h.vn/hoi-dap/question/821353.html
\(B=1-\frac{1}{2.5}+1-\frac{1}{5.8}+...+1-\frac{1}{50.53}\)
\(B=17-\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+....+\frac{3}{50.53}\right)\)
\(B=17-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-.....-\frac{1}{53}\right)\)
\(B=17-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{53}\right)=17-\frac{1}{3}.\frac{51}{106}=17-\frac{17}{106}=17\left(1-\frac{1}{106}\right)=17.\frac{105}{106}=\frac{1785}{106}\)