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29 tháng 6 2022
\(=\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{29\cdot32}\)
\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{29}-\dfrac{1}{32}\)
=1/2-1/32=15/32
LA
2
24 tháng 4 2015
a)\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=\left(1-\frac{1}{6}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+...+\left(\frac{1}{5}-\frac{1}{5}\right)\)
\(=\left(1-\frac{1}{6}\right)+0+...+0=1-\frac{1}{6}=\frac{6}{6}-\frac{1}{6}=\frac{5}{6}\)
b)\(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)
\(=\left(\frac{1}{2}-\frac{1}{14}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+\left(\frac{1}{11}-\frac{1}{11}\right)\)
\(=\left(\frac{1}{2}-\frac{1}{14}\right)+0+...+0=\frac{1}{2}-\frac{1}{14}=\frac{7}{14}-\frac{1}{14}=\frac{6}{14}\)
Nhớ **** cho mình nhé bạn! chúc bạn học tốt
NN
1
N
2
16 tháng 5 2018
\(A=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)
\(3A=3.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\right)\)
\(3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\)
\(3A=\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+\frac{17-14}{14.17}+\frac{20-17}{17.20}\)
\(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)
\(3A=\frac{1}{2}-\frac{1}{20}\)
\(A=\left(\frac{1}{2}-\frac{1}{20}\right)\div3=\frac{9}{20}\div3=\frac{9}{20.3}=\frac{3}{20}\)
Vậy ................
\(B=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot....\cdot\frac{9999}{10000}\)
\(B=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot...\cdot\frac{99.101}{100.100}\)
\(B=\frac{\left(1\cdot2\cdot3\cdot...\cdot99\right).\left(3\cdot4\cdot5\cdot...\cdot101\right)}{\left(2\cdot3\cdot4\cdot...\cdot100\right).\left(2\cdot3\cdot4\cdot...\cdot100\right)}\)
\(B=\frac{1\cdot2\cdot3\cdot..\cdot99}{2\cdot3\cdot4\cdot..\cdot100}\cdot\frac{3\cdot4\cdot5\cdot...\cdot101}{2\cdot3\cdot4\cdot...\cdot100}\)
\(B=\frac{1}{100}\cdot\frac{101}{2}=\frac{101}{200}\)
vậy......
16 tháng 5 2018
A=1/2.5+1/5.8+1/8.11+1/11.14+1/14.17+1/17.20
A=1/3.(3/2.5+3/5.8+3/8.11+3/11.14+3/14.17+3/17.20)
A=1/3.(1/2-1/20)
=3/20
B=1.3/2.2+2.4/3.3+3.5/4.4+...+99.101/100.100
B=(1.2.3...99).(3.4.5...101)/(2.3.4...100).(2.3.4...100)
B=\(\frac{1.2....99}{2.3...100}\).\(\frac{3.4...101}{2.3...100}\)
B=1/100.101/2=101/200
30 tháng 4 2018
\(\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(3\left(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{3\cdot101}{1540}\)
\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{308}{1540}-\frac{303}{1540}=\frac{5}{1540}=\frac{1}{308}\)
\(x+3=308\)
\(x=308-3=305\)
DD
25 tháng 4 2021
\(\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+...+\dfrac{1}{y.(y+3)}=\dfrac{98}{1545}\)
\(3.(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{y(y+3)})=\dfrac{98}{1545}.3\)
\(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{y(y+3)}=\dfrac{98}{515}\)
\(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{y}-\dfrac{1}{y+3}=\dfrac{98}{515}\)
\(\dfrac{1}{y+3}=\dfrac{98}{515}\)
\(y+3.1=1.515\)
\(y+3=515\)
\(y=515-3\)
\(y=512\)
Vậy y = 512
Nhớ tick cho mk nha
9 tháng 7 2020
\(\frac{2}{40}+\frac{2}{88}+\frac{2}{154}+...+\frac{2}{x\left(x+3\right)}=\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+...+\frac{2}{x\left(x+3\right)}\)
\(=\frac{2}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{2}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{x}-\frac{1}{x+3}\right)\)
\(=\frac{2}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)\)
Từ đó ta có:
\(\frac{2}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{202}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
\(x+3=308\)
x = 305
ND
0
3/7 nha bạn
\(\frac{3}{10}\)+\(\frac{3}{40}\)+\(\frac{3}{88}\)+\(\frac{3}{154}\)
=\(\frac{3}{2.5}\)+\(\frac{3}{5.8}\)+\(\frac{3}{8.11}\)+\(\frac{3}{11.14}\)
Đến đây bạn tự suy luận tiếp nhé,mình không viết vào vì nó không cần thiết,phân tích xong nó sẽ như thế này:
=\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)
Loại bỏ những phân số đối nhau ta còn:
=\(\frac{1}{2}-\frac{1}{14}\)
=\(\frac{7}{14}-\frac{1}{14}\)
=\(\frac{3}{7}\)
Chúc bạn học tốt^^