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\(\dfrac{10}{11}:\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}\right)\)
\(=\dfrac{10}{11}:\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)\)
\(=\dfrac{10}{11}:\left(\dfrac{1}{3}-\dfrac{1}{11}\right)\)
\(=\dfrac{10}{11}:\dfrac{8}{33}\)
\(=\dfrac{10}{11}\times\dfrac{33}{8}\)
\(=5\times\dfrac{3}{4}\)
\(=\dfrac{15}{4}\)
Đây là tổng của 2 dãy:
\(\frac{1}{1\times3\times5}+\frac{1}{3\times5\times7}+\frac{1}{5\times7\times9}+...+\frac{1}{995\times997\times999}\)(1)
và
\(\frac{1}{2\times5\times8}+\frac{1}{5\times8\times11}+\frac{1}{8\times11\times14}+...+\frac{1}{1493\times1496\times1499}\)(2)
Dãy số có dạng là tích 3 thừa số, trong đó thừa số thứ 3 hơn thừa số thứ nhất n đơn vị và 2 thừa số cuối của phân số trước là 2 thừa số đầu của phân số sau. Để tính dãy kiểu này cần đưa tử số về hiệu của thừa số thứ 3 và thừa số thứ nhất (hiệu = n):
Vậy nhân dãy thứ nhất với 4:
\(=\frac{4}{1\times3\times5}+\frac{4}{3\times5\times7}+\frac{4}{5\times7\times9}+...+\frac{4}{995\times997\times999}\)
Nhận xét:
- \(\frac{4}{1\times3\times5}=\frac{5-1}{1\times3\times5}=\frac{5}{1\times3\times5}-\frac{1}{1\times3\times5}=\frac{1}{1\times3}-\frac{1}{3\times5}\)
- \(\frac{4}{3\times5\times7}=\frac{7-3}{3\times5\times7}=\frac{7}{3\times5\times7}-\frac{3}{3\times5\times7}=\frac{1}{3\times5}-\frac{1}{5\times7}\)
Vậy 4 lần tổng dãy 1 là:
\(\frac{1}{1\times3}-\frac{1}{3\times5}+\frac{1}{3\times5}-\frac{1}{5\times7}+...+\frac{1}{995\times997}-\frac{1}{997\times999}\)
\(\frac{1}{1\times3}-\frac{1}{997\times999}\)
Suy ra tổng dãy (1) là \(\left(\frac{1}{3}-\frac{1}{997\times999}\right)\times\frac{1}{4}\)
Làm tương tự tính được tổng dãy (2) là: \(\left(\frac{1}{2\times5}-\frac{1}{1496\times1499}\right)\times\frac{1}{6}\)
Cộng 2 kết quả lại được tổng cần tính
\(1,\\ =\dfrac{2-1}{1\times2}+\dfrac{3-2}{2\times3}+\dfrac{4-3}{3\times4}+\dfrac{5-4}{4\times5}+.....+\dfrac{99-98}{98\times99}+\dfrac{100-99}{99\times100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{100-1}{100}=\dfrac{99}{100}\)
\(2,=\dfrac{13-11}{11\times13}+\dfrac{15-13}{13\times15}+....+\dfrac{21-19}{19\times21}+\dfrac{23-21}{21\times23}\\ =\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+....+\dfrac{1}{19}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{23}\\ =\dfrac{1}{11}-\dfrac{1}{23}\\ =\dfrac{23-11}{11\times23}=\dfrac{12}{253}\)
@seven
a: 1/1*2+1/2*3+...+1/99*100
=1-1/2+1/2-1/3+...+1/99-1/100
=1-1/100
=99/100
b: 2/11*13+2/13*15+...+2/21*23
=1/11-1/13+1/13-1/15+...+1/21-1/23
=1/11-1/23
=12/253
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
\(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+...+\dfrac{2}{48\times49\times50}\)
\(=\dfrac{1}{1\times2}-\dfrac{1}{2\times3}+\dfrac{1}{2\times3}-\dfrac{1}{3\times4}+\dfrac{1}{3\times4}-\dfrac{1}{4\times5}+...+\dfrac{1}{48\times49}-\dfrac{1}{49\times50}\)
\(=\dfrac{1}{1\times2}-\dfrac{1}{49\times50}\)
\(=\dfrac{1}{2}-\dfrac{1}{2450}\)
\(=\dfrac{612}{1225}\)
\(\text{#}Toru\)
\(C=\dfrac{1}{8}\times\dfrac{3}{2}-\dfrac{1}{8}\times\dfrac{1}{4}+\dfrac{3}{8}\times\dfrac{5}{8}\)
\(C=\dfrac{1}{8}\times\left(\dfrac{3}{2}-\dfrac{1}{4}\right)+\dfrac{3}{8}\times\dfrac{5}{4}\)
\(C=\dfrac{1}{8}\times\dfrac{5}{4}+\dfrac{3}{8}\times\dfrac{5}{4}\)
\(C=\left(\dfrac{1}{8}+\dfrac{3}{8}\right)\times\dfrac{5}{4}\)
\(C=\dfrac{1}{2}\times\dfrac{5}{4}\)
\(C=\dfrac{5}{8}\)
Giải:
\(\dfrac{1}{2}+\dfrac{2}{8}+\dfrac{3}{28}+\dfrac{4}{77}+\dfrac{5}{176}+\dfrac{6}{352}\)
\(=\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+\dfrac{4}{7.11}+\dfrac{5}{11.16}+\dfrac{6}{16.22}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{22}\)
\(=\dfrac{1}{1}-\dfrac{1}{22}\)
\(=\dfrac{21}{22}\)
\(\dfrac{1}{2}+\dfrac{2}{8}+\dfrac{3}{28}+\dfrac{4}{77}+\dfrac{5}{176}+\dfrac{6}{352}\\ =\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{28}+\dfrac{4}{77}+\dfrac{5}{176}+\dfrac{3}{176}\\ =\dfrac{3}{4}+\dfrac{3}{28}+\dfrac{4}{77}+\dfrac{1}{22}\\ =\dfrac{21}{28}+\dfrac{3}{28}+\dfrac{7}{154}+\dfrac{8}{154}\\ =\dfrac{6}{7}+\dfrac{15}{154}\\ =\dfrac{21}{22}\)
1)\(y\times7:5+4\times8=134\)
\(\Leftrightarrow y\times7:5+32=134\)
\(\Leftrightarrow y\times7:5=102\)
\(\Leftrightarrow y\times7=510\)
\(\Leftrightarrow y=72,86\)
2) \(\dfrac{1}{4}:0,25-\dfrac{1}{8}:0,125+\dfrac{1}{2}:0,5-\dfrac{1}{10}\)
\(=0,25:0,25-0,125:0,125+0,5:0,5-\dfrac{1}{10}\)
\(=1-1+1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
\(\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{3}{3}=1\)
\(\dfrac{4}{5}+\dfrac{5}{6}=\dfrac{24}{30}+\dfrac{25}{30}=\dfrac{49}{30}\)
\(\dfrac{4}{5}-\dfrac{3}{5}=\dfrac{1}{5}\)
\(\dfrac{8}{5}x\dfrac{5}{8}=\dfrac{1}{1}=1\)
\(\dfrac{6}{7}x\dfrac{4}{7}=\dfrac{24}{49}\)
\(\dfrac{4}{5}:\dfrac{4}{5}=\dfrac{4}{5}x\dfrac{5}{4}=\dfrac{1}{1}=1\)
\(\dfrac{5}{5}:\dfrac{5}{5}=\dfrac{5}{5}x\dfrac{5}{5}=\dfrac{1}{1}=1\)
1) \(\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{1+2}{3}=\dfrac{3}{3}=1\)
2) \(\dfrac{4}{5}+\dfrac{5}{6}=\dfrac{24}{30}+\dfrac{25}{30}=\dfrac{24+25}{30}=\dfrac{49}{30}\)
3) \(\dfrac{4}{5}-\dfrac{3}{5}=\dfrac{4-3}{5}=\dfrac{1}{5}\)
4) \(\dfrac{9}{8}-\dfrac{4}{2}=\dfrac{9}{8}-2=\dfrac{9}{8}-\dfrac{16}{8}=-\dfrac{7}{8}\)
5) \(\dfrac{8}{5}\times\dfrac{5}{8}=\dfrac{8\times5}{5\times8}=\dfrac{40}{40}=1\)
6) \(\dfrac{6}{7}\times\dfrac{4}{7}=\dfrac{6\times4}{7}=\dfrac{24}{7}\)
7) \(\dfrac{4}{5}:\dfrac{4}{5}=\dfrac{4}{5}\times\dfrac{5}{4}=\dfrac{4\times5}{5\times4}=\dfrac{20}{20}=1\)
8) \(\dfrac{5}{5}:\dfrac{5}{5}=\dfrac{5}{5}\times\dfrac{5}{5}=\dfrac{5\times5}{5\times5}=\dfrac{25}{25}=1\)
`25%+3/4+1/2:0,5-1/4:0,15+1/8:0,125`
`=1/4+3/4+1/2xx2-1/4xx4+1/8xx8`
`(1/4+3/4)+(1/2xx2)-(1/4xx4)+(1/8xx8)`
`=1+1-1+1`
`=2`
A = \(\dfrac{8}{1\times3}\) + \(\dfrac{8}{3\times5}\) + \(\dfrac{8}{5\times7}\) +......+\(\dfrac{8}{99\times101}\)
A = 4 x ( \(\dfrac{2}{1\times3}\) + \(\dfrac{2}{3\times5}\) + \(\dfrac{2}{5\times7}\) +........+ \(\dfrac{2}{99\times101}\)
A = 4 x ( \(\dfrac{1}{1}\) - \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\)- \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) +........+ \(\dfrac{1}{99}\) - \(\dfrac{1}{101}\))
A = 4 x ( 1 - \(\dfrac{1}{101}\) )
A = \(\dfrac{400}{101}\)
\(\dfrac{8}{1\times3}+\dfrac{8}{3\times5}+\dfrac{8}{5\times7}+...+\dfrac{8}{99\times101}\)
\(=4\left(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{99\times101}\right)\)
\(=4\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=4\left(\dfrac{1}{1}-\dfrac{1}{101}\right)\)
\(=4.\dfrac{100}{101}\)
\(=\dfrac{400}{101}\)