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\(\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right).\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)\)
\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right).\left(\frac{-5}{12}+\frac{3}{12}+\frac{2}{12}\right)\)
\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right).\left(\frac{-2}{12}+\frac{2}{12}\right)\)
\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right).0\)
\(=0\)
Ta có: A = \(\frac{6}{5\times7}+\frac{6}{7\times9}+\frac{6}{9\times11}+...+\frac{6}{95\times97}+\frac{6}{97\times99}\)
\(\Rightarrow A=\frac{1}{6}\left(\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+...+\frac{1}{95\times97}+\frac{1}{97\times99}\right)\)
\(\Rightarrow A=\frac{1}{6}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Rightarrow A=\frac{1}{6}\left(\frac{1}{5}-\frac{1}{99}\right)\)
=> A = ...
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
=（2-1）*（2+1）+（4-1）*（4+1）+ ...+(2n-1)*(2n+1) =(2^2-1)+(4^2-1)+...+(4n^2-1) =(2^2+4^2+...+4n^2)-(1+1+...+1) =4(1^2+2^2+...n^2)-n n(n+1)(2n+1)/6： 1^2+2^2+3^2+…+n^2=n(n+1)(2n+1)/6n^2=n 1x3+3x5+5x7+7x9+...+17x19 =4(1^2+2^2+...n^2)-n =4*n(n+1)(2n+1)/6-n； n=10，1x3+3x5+5x7+7x9+...+17x19=1530
b)
S2=6/2x5+6/5x8+6/8x11+...+6/29x32
=2.(3/2.5+3/5.8+...+3/29.32)
=2.(1/2-1/5+1/5-1/8+...+1/29-1/32)
=2.(1/2-1/32)
=2.15/32
=15/16
a)
Ta có:
S1=2/3x5+2/5x7+2/7x9+...+2/97x99
=1/3-1/5+1/5-1/7+...+1/97-1/99
=1/3-1/99
=32/99
Câu 11:
(\(\dfrac{11}{4}\). \(\dfrac{-5}{9}\) - \(\dfrac{4}{9}\).\(\dfrac{11}{4}\)).\(\dfrac{8}{33}\)
= \(\dfrac{11}{4}\).(\(\dfrac{-5}{9}\) - \(\dfrac{4}{9}\)). \(\dfrac{8}{33}\)
= \(\dfrac{11}{4}\).(-1).\(\dfrac{8}{33}\)
= - \(\dfrac{2}{3}\)
a) \(a=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{37\cdot39}\)
\(a=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)
\(a=\frac{1}{3}-\frac{1}{39}\)
\(a=\frac{12}{39}\)
b) \(\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)\cdot\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)\)
\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)\cdot\left(\frac{-5}{12}+\frac{3}{12}+\frac{2}{12}\right)\)
\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)\cdot\left(\frac{-2}{12}+\frac{2}{12}\right)\)
\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)\cdot0\)
\(=0\)
a) \(A=\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+...+\frac{2}{37x39}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{37}-\frac{1}{39}\)
\(A=\frac{1}{3}-\frac{1}{39}\)
\(A=\frac{4}{13}\)
b) \(\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)x\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)\)
\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)x0\)
\(=0\)