Tính tổng S = 10.(\(\frac{1}{2.12}\)+\(\frac{1}{12.22}\)+\(\frac{1}{22.32}\)+\(\frac{1}{32.42}\)+.........+\(\frac{1}{2002.2012}\))
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\(S=\frac{1}{2\cdot12}+\frac{1}{12\cdot22}+...+\frac{1}{2002\cdot2012}\)
\(S=\left(\frac{1}{2\cdot12}+\frac{1}{12\cdot22}...+\frac{1}{2002\cdot2012}\right)\cdot\frac{1}{10}\)
\(S=\left(\frac{1}{2}-\frac{1}{12}+\frac{1}{12}-\frac{1}{22}+...+\frac{1}{2002}-\frac{1}{2012}\right)\cdot\frac{1}{10}\)
\(S=\left(\frac{1}{2}-\frac{1}{2012}\right)\cdot\frac{1}{10}\)
\(S=\frac{1006-1}{2012}\cdot\frac{1}{10}\)
\(S=\frac{1005}{2012}\cdot\frac{1}{10}\)
\(S=\frac{201}{2012}\cdot\frac{1}{2}\)
\(S=\frac{201}{4024}\)
S=1/10 \(\times\)( 1/2 - 1/12 + 1/12 - 1/22 +...+1/2002 -1/2012 )
= 1/10 \(\times\)( 1/2 - 1/2012)
= 1/10 \(\times\)1005/2012
= 201/4024
Bài 1 :
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(S=\frac{1}{1}-\frac{1}{2011}=\frac{2010}{2011}\)
Bài 2 :
\(S=\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+...+\frac{1}{58}-\frac{1}{61}\)
\(S=\frac{1}{10}-\frac{1}{61}=\frac{51}{610}\)
Bài 3 :
\(3S=\frac{3}{4\times7}+\frac{3}{7\times11}+...+\frac{3}{19\times22}\)
\(3S=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{19}-\frac{1}{22}\)
\(3S=\frac{1}{4}-\frac{1}{22}\)
\(S=\frac{18}{88}\div3=\frac{6}{88}\)
cau a),b),c) ban dat mau chung roi khu mau ma lam la duoc ma
Tuy học lớp 6 ................. cơ mừ thấy mí bài nỳ dễ quá >.<
đề sai thì phải
\(A=\frac{10}{2\cdot12}+\frac{2}{3\cdot5}+\frac{3}{5\cdot8}+\frac{1}{2\cdot3}+\frac{5}{12\cdot17}+\frac{6}{17\cdot23}+\frac{7}{23\cdot30}\)
\(A=\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{2}-\frac{1}{3}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\)
\(A=\frac{1}{2}+\frac{1}{2}-\frac{1}{8}-\frac{1}{30}\)
\(A=\frac{101}{120}\)
\(A=\frac{1}{2.12}+\frac{2}{3.5}+\frac{3}{5.8}+...+\frac{7}{23.30}\)
\(=\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\)
\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{8}-\frac{1}{30}=1-\frac{19}{120}=\frac{101}{120}\)
\(100E\)\(=\frac{100}{1.101}+\frac{100}{2.102}+..........+\frac{100}{10.110}\)
\(=1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+........+\frac{1}{10}-\frac{1}{110}\)
\(10F=\frac{10}{1.11}+\frac{10}{2.12}+......+\frac{10}{100.110}\)
\(=1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+......+\frac{1}{100}-\frac{1}{110}\)
\(=1+\frac{1}{2}+...+\frac{1}{10}+\frac{1}{11}+....+\frac{1}{100}-\frac{1}{11}-\frac{1}{12}-....-\frac{1}{100}-\frac{1}{101}-...-\frac{1}{110}\)
\(=1+\frac{1}{2}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\)\(=100E\)
\(\Rightarrow10F=100E\Rightarrow\frac{E}{F}=\frac{1}{10}\)
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