TÍNH
\(\frac{5}{3.5}+\frac{5}{5.7}+\frac{5}{7.9}+....+\frac{5}{95.97}\)
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\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
Tự tính
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
\(S=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)
S=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+......+\frac{1}{95.97}+\frac{1}{97.99}\)
S=\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{97}-\frac{1}{99}\right)\)
S=\(\frac{1}{2}.\left(1-\frac{1}{99}\right)\)
S=\(\frac{1}{2}.\frac{98}{99}\)
S=\(\frac{49}{99}\)
a) \(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+.....+\frac{5}{27.30}\)
\(=\frac{5}{3}\left(\frac{1}{1.4}+\frac{1}{4.7}+........+\frac{1}{27.30}\right)\)
\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{27}-\frac{1}{30}\right)\)
\(=\frac{5}{3}\left(1-\frac{1}{30}\right)\)
\(=\frac{5}{3}.\frac{29}{30}=\frac{29}{36}\)
Đặt \(A=\frac{12}{3\cdot5}+\frac{12}{5\cdot7}+\frac{12}{7\cdot9}+....+\frac{12}{97\cdot99}\)
\(2A=\frac{12}{3}-\frac{12}{5}+\frac{12}{5}-\frac{12}{7}+...+\frac{12}{97}-\frac{12}{99}\)
\(2A=\frac{12}{3}-\frac{12}{99}\)
\(A=\frac{128}{33}\cdot\frac{1}{2}=\frac{64}{33}\)
S = \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+\frac{5}{7.9}+.......+\frac{5}{17.19}\)
S : 5 = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{17.19}\)
S : 5 = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}+.......+\frac{1}{17}-\frac{1}{19}\)
=> S : 5 = 1 - \(\frac{1}{19}=\frac{19}{19}-\frac{1}{19}=\frac{18}{19}\)
=> S = \(\frac{18}{19}x5=\frac{90}{19}\)
a, Ta có \(A=\frac{3}{3.5}+\frac{3}{5.7}+....+\frac{3}{49.51}\)
\(=\frac{3}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{49.51}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)\)
\(=\frac{1}{2}-\frac{3}{102}=\frac{48}{102}=\frac{24}{51}\)
b,Ta có \(\frac{1}{2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}\)
\(=\frac{2-1}{2}+\frac{4-2}{2.4}+\frac{7-4}{4.7}+\frac{11-7}{7.11}+\frac{16-11}{11.16}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}\)
\(=\frac{15}{16}\)
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!!!!!!!!!!!!!!!!!!!
!!!!!!!!!!!!!!1111
\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{n\left(n+2\right)}=\frac{5}{36}\)
\(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{n\left(n+2\right)}\right)=\frac{5}{36}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+2}=\frac{5}{18}\)
\(\frac{1}{3}-\frac{1}{n+2}=\frac{5}{18}\)
\(\frac{1}{n+2}=\frac{1}{18}\)
\(\Rightarrow n+2=18\Rightarrow n=16\)
\(\Rightarrow\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{n.\left(n+2\right)}=\frac{10}{36}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+2}=\frac{5}{18}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{n+2}=\frac{5}{18}\)
\(\Rightarrow\frac{n+2-3}{3\left(n+2\right)}=\frac{5}{18}\)
\(\Rightarrow\frac{n-1}{3n+6}=\frac{5}{18}\)
\(\Rightarrow18\left(n-1\right)=5\left(3n+6\right)\)
\(\Rightarrow18n-18=15n+30\)
\(\Rightarrow3n=48\)
\(\Rightarrow n=48:3\)
=>n=16
=(1/3-1/5)+(1/5-1/7)=.........+1/95-1/97
=1/3-1/97
=94/291
ok