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Đặt : \(A=\frac{5}{1\cdot4}+\frac{5}{4\cdot7}+\frac{5}{7\cdot10}+...+\frac{5}{27\cdot30}\)
\(A=\frac{1}{3}\left(\frac{5}{1}-\frac{5}{4}+\frac{5}{4}-\frac{5}{7}+...+\frac{5}{27}-\frac{5}{30}\right)\)
\(A=\frac{1}{3}\left(5-\frac{5}{30}\right)\)
\(A=\frac{1}{3}\cdot\frac{29}{6}\)
\(A=\frac{29}{18}\)
\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+....+\frac{5}{27.30}\)
\(=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{30-27}{27.30}\)
\(=\frac{5}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{27}-\frac{1}{30}\right)\)
\(=\frac{5}{3}\cdot\left(1-\frac{1}{30}\right)\)
\(=\frac{5}{3}\cdot\frac{29}{30}=\frac{29}{18}\)
1/3xD=1/(2x4)+1/(4x6)+...+1/(98x100)
2/3xD=2/(2x4)+2/(4x6)+...+1/(98x100)
2/3xD= 1/2-1/4+1/4-1/6+...+1/98-1/100
2/3xD=1/2-1/100
2/3xD=49/100
D=147/200
a) 2/9 +1/5 +7/9+4/5
=( 2/9+7/9)+(1/5+4/5)
=1+1=2
b) 1/12+3/16+5/12+5/16
=(1/12 +5/12)+(3/16+5/16)
=1/2 +1/2=1
\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)
=\(\frac{3}{1.2}+\frac{3}{2.4}+\frac{3}{4.8}+\frac{3}{8.16}+\frac{3}{16.32}\)
=\(\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{4}+\frac{3}{4}-\frac{3}{8}+\frac{3}{8}-\frac{3}{16}+\frac{3}{16}-\frac{3}{36}\)
=\(\frac{3}{1}-\frac{3}{36}\)=\(\frac{35}{12}\)
b
Q=\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{9900}\)
Rồi giải tương tự như câu a là được
M=\(5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=5\left(1-\frac{1}{100}\right)=5.\frac{99}{100}=\frac{99}{20}\)
a) \(\frac{2}{5}+\left(\frac{2}{8}+\frac{3}{5}+\frac{6}{8}\right)=\left(\frac{2}{5}+\frac{3}{5}\right)+\left(\frac{2}{8}+\frac{6}{8}\right)=1+1=2\)
b) \(\frac{6}{9}+\left(\frac{15}{25}+\frac{12}{9}+\frac{6}{15}\right)=\left(\frac{6}{9}+\frac{12}{9}\right)+\left(\frac{15}{25}+\frac{6}{15}\right)\)
\(=2+\left(\frac{3}{5}+\frac{2}{5}\right)=2+1=3\)
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 \(A=\frac{12}{3\cdot5}+\frac{12}{5\cdot7}+\frac{12}{7\cdot9}+....+\frac{12}{97\cdot99}\)
\(2A=\frac{12}{3}-\frac{12}{5}+\frac{12}{5}-\frac{12}{7}+...+\frac{12}{97}-\frac{12}{99}\)
\(2A=\frac{12}{3}-\frac{12}{99}\)
\(A=\frac{128}{33}\cdot\frac{1}{2}=\frac{64}{33}\)