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A= \(\frac{-1}{4\cdot5}+\frac{-1}{5\cdot6}+\frac{-1}{6\cdot7}+\frac{-1}{7\cdot8}+\frac{-1}{8\cdot9}+\frac{-1}{9\cdot10}\)
=\(-1\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\right)\)
=\(-1\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\right)\)
=\(-1\left(\frac{1}{4}-\frac{1}{10}\right)\)
=\(-1\cdot\frac{3}{20}\)
=\(\frac{-3}{20}\)
=\(\frac{-1}{20}\)
\(M=\left(\dfrac{3}{2}-\dfrac{5}{6}+\dfrac{7}{12}-\dfrac{17}{72}\right)+\left(-\dfrac{9}{20}+\dfrac{11}{30}\right)+\left(\dfrac{-13}{42}+\dfrac{15}{56}\right)\)
\(=\dfrac{108}{72}-\dfrac{60}{72}+\dfrac{42}{72}-\dfrac{17}{72}+\dfrac{-27}{60}+\dfrac{22}{60}+\dfrac{-52}{168}+\dfrac{45}{168}\)
\(=\dfrac{73}{72}-\dfrac{1}{12}-\dfrac{1}{24}=\dfrac{73}{72}-\dfrac{6}{72}-\dfrac{3}{72}=\dfrac{64}{72}=\dfrac{8}{9}\)
a) \(\frac{1}{n}\) - \(\frac{1}{n+1}\) = \(\frac{n+1}{n\left(n+1\right)}\) - \(\frac{n}{n\left(n+1\right)}\) = \(\frac{1}{n\left(n+1\right)}\) = \(\frac{1}{n}\) . \(\frac{1}{n+1}\) =>đpcm
b) A= \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\)+...+\(\frac{1}{8}\) - \(\frac{1}{9}\) +\(\frac{1}{9}\)
= \(\frac{1}{2}\) + \(\frac{1}{9}\)= \(\frac{11}{18}\)
\(=\left(1+\frac{1}{2}\right)-1+\frac{1}{6}+\left(\frac{1}{2}+\frac{1}{12}\right)-\frac{1}{2}+\frac{1}{20}+\left(\frac{1}{3}+\frac{1}{30}\right)-\frac{1}{3}+\frac{1}{42}+\left(\frac{1}{4}+\frac{1}{56}\right)-\frac{1}{4}+\frac{1}{72}\)
=\(=\left(1-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{72}\right)\)
\(=0+\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}\right)=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}=\left(1-\frac{1}{9}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+...+\left(\frac{1}{8}-\frac{1}{8}\right)\)\(=\left(\frac{9}{9}-\frac{1}{9}\right)+0+...+0=\frac{8}{9}\)
\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow1-\frac{1}{100}=\frac{99}{100}\)
Vậy B = \(\frac{99}{100}\)
S=10/2.12+10/12.22+10/22.32+10/32.42+.......+10/2002.2012
S=1/2-1/12+1/12-1/22+1/22-1/32+1/32-1/42+.....+1/2002-1/2012
S=1/2-1/2012
S=????
bạn tự tính nhé
S=10.1/10{1/2-1/12+1/12-1/22+1/22-1/32+...+1/2002-1/2012}
=1/2-1/2012
=1005/2012
\(\frac{30}{43}=\frac{1}{\frac{43}{30}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{1+\frac{1}{\frac{30}{13}}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{1+\frac{1}{2+\frac{1}{\frac{13}{4}}}}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)
=> a = 1; b = 2; c = 3; d = 4.
a) ta có:
\(\frac{-1}{2}-1\le x\le\frac{1}{2}.3\)
hay \(-1,5\le x\le1,5\)
vì x\(\in Z\) nên ta chọn x=-1,0,1
ta có:
3S=\(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)
3S-S=\(\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^9}\right)\)
2S=1-\(\frac{1}{3^9}\)
s=\(\left(1-\frac{1}{3^9}\right):2\)
M=1/2{1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99}
=1/2{1/3-1/99}
=1/2*32/99
=16/99
\(A=-\frac{1}{20}+-\frac{1}{30}+...+-\frac{1}{90}\)
\(=-\frac{1}{4.5}+-\frac{1}{5.6}+...+-\frac{1}{9.10}\)
\(=\left(-\frac{1}{4}\right)-\left(-\frac{1}{5}\right)+\left(-\frac{1}{5}\right)-\left(-\frac{1}{6}\right)+...+\left(-\frac{1}{9}\right)-\left(-\frac{1}{10}\right)\)
\(=\left(-\frac{1}{4}\right)-\left(-\frac{1}{10}\right)=-\frac{3}{20}\)
Vậy \(A=-\frac{3}{20}\)