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\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
\(=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+\frac{1}{14\cdot17}+\frac{1}{17\cdot20}\)
\(=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+\frac{3}{14\cdot17}+\frac{3}{17\cdot20}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(=\frac{1}{3}\cdot\frac{9}{20}\)
\(=\frac{3}{20}\)
\(A=\frac{3}{3}.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\right)\)
\(A=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{3}{20}\)
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)
\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{3}{20}\)
S=1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 + 1/14.17 + 11/17.20
S=1/3.(1/2-1/5+1/5-1/8+.......+1/17 - 1/20)
S=1/3.(1/2-1/20)
S=1/3.9/20
S=3/20
Ai tích mk mk sẽ tích lại
Chú ý tích gấp ddooi khi tích đủ 3 cái
Ta có: S = 1/10 + 1/40 + 1/88 + 1/154 + 1/238 + 1/340 + 1/460 + 1/598 + 1/754 + 1/928
=> S = 1/2.5 + 1/5.8 + 1/8.11 + ... + 1/26.29 + 1/29.32
Nhân 2 vế với 3 và áp dụng công thức tách 1 phân số thành hiệu 2 phân số: x/n.(n + x) = 1/n - 1/(n + x)
=> 3.S = 3.(1/2.5 + 1/5.8 + 1/8.11 + ... + 1/26.29 +1/29.32)
=> 3.S = 3/2.5 + 3/5.8 + 3/8.11 + ... + 3/26.29 + 3/29.32
=> 3.S = 1/2 - 1/ 5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/26 - 1/29 + 1/29 - 1/32
=> 3.S = 1/2 - 1/32
=> 3.S = 15/32
=> S = 5/32
@Cre: G+
S=1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 + 1/14.17 + 11/17.20
S=1/3.(1/2-1/5+1/5-1/8+.......+1/17 - 1/20)
S=1/3.(1/2-1/20)
S=1/3.9/20
S=3/20
\(A=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)
\(3A=3.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\right)\)
\(3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\)
\(3A=\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+\frac{17-14}{14.17}+\frac{20-17}{17.20}\)
\(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)
\(3A=\frac{1}{2}-\frac{1}{20}\)
\(A=\left(\frac{1}{2}-\frac{1}{20}\right)\div3=\frac{9}{20}\div3=\frac{9}{20.3}=\frac{3}{20}\)
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......
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
S=\(\frac{1}{10}\)+ \(\frac{1}{40}\)+\(\frac{1}{88}\)+\(\frac{1}{154}\)+\(\frac{1}{238}\)+\(\frac{1}{340}\)
S=\(\frac{1}{2.5}\)+\(\frac{1}{5.8}\)+\(\frac{1}{8.11}\)+\(\frac{1}{11.14}\)+\(\frac{1}{14.17}\)+\(\frac{1}{17.20}\)
S= \(\frac{1}{3}\).(\(\frac{3}{2.5}\)+\(\frac{3}{5.8}\)+\(\frac{3}{8.11}\)+\(\frac{3}{11.14}\)+\(\frac{3}{14.17}\)+\(\frac{3}{17.20}\))
S= \(\frac{1}{3}\).(\(\frac{1}{2}\)-\(\frac{1}{20}\))
S= \(\frac{1}{3}\).\(\frac{9}{20}\)
S=\(\frac{3}{20}\)
\(\frac{1}{10}\) + \(\frac{1}{40}\) +\(\frac{1}{88}\)+\(\frac{1}{154}\) + \(\frac{1}{238}\) + \(\frac{1}{340}\)
= \(\frac{1}{8}\) + \(\frac{1}{56}\) + \(\frac{1}{140}\)
= \(\frac{1}{7}\) + \(\frac{1}{140}\)
= \(\frac{3}{20}\)
\(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+\dfrac{1}{238}\)
\(=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+\dfrac{1}{14.17}\)
\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{14}-\dfrac{1}{17}\)
\(=\dfrac{1}{2}-\dfrac{1}{17}\)
\(=\dfrac{17}{34}-\dfrac{2}{34}\)
\(=\dfrac{15}{34}\)
= 5/ 34