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Ta có: \(B=\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\)
\(\Rightarrow B=\frac{2}{20}+\frac{2}{30}+...+\frac{2}{240}\)
\(\Rightarrow B=2.\left(\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\right)\)
\(\Rightarrow B=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{15.16}\right)\)
\(\Rightarrow B=2.\left(\frac{1}{4}-\frac{1}{16}\right)=2.\frac{3}{16}=\frac{3}{8}\)
Vậy \(B=\frac{3}{8}\)
nha m.n
\(B=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+.....+\frac{1}{120}\)
\(B=2.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+.....+\frac{1}{240}\right)\)
\(B=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{15.16}\right)\)
\(B=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+......+\frac{1}{15}-\frac{1}{16}\right)\)
\(B=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(B=2.\frac{3}{16}\)
\(B=\frac{3}{8}\)
Vậy \(B=\frac{3}{8}\)

\(B=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)
\(\Leftrightarrow B=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)
\(\Leftrightarrow B=2\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{240}\right)\)
\(\Leftrightarrow B=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)
\(\Leftrightarrow B=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(\Leftrightarrow B=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{3}{8}\)
Vì \(\dfrac{3}{8}< \dfrac{1}{2}\)
\(\Rightarrow B< \dfrac{1}{2}\left(ĐPCM\right)\)

B = \(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+....+\frac{1}{120}\)
\(B=\frac{1}{4.5:2}+\frac{1}{5.6:2}+\frac{1}{6.7:2}+.....+\frac{1}{15.16:2}\)
\(\frac{1}{2}B=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{15.16}\)
Ta thấy: \(\frac{1}{4.5}=\frac{1}{4}-\frac{1}{5};\frac{1}{5.6}=\frac{1}{5}-\frac{1}{6};\frac{1}{6.7}=\frac{1}{6}-\frac{1}{7};.....\)
\(\frac{1}{2}B=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-.....-\frac{1}{15}+\frac{1}{15}-\frac{1}{16}\)
\(\frac{1}{2}B=\frac{1}{4}-\frac{1}{16}=\frac{3}{16}\)
\(B=\frac{3}{16}:\frac{1}{2}=\frac{3}{16}.2=\frac{3}{8}\)

\(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)
\(=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)
\(=2\times\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...\dfrac{1}{240}\right)\)
\(=2\times\left(\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+...+\dfrac{1}{15\times16}\right)\)
\(=2\times\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(=2\times\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)
\(=\dfrac{3}{8}\)
=2/20+2/30+2/42+.....+2/240
=2/4.5+2/5.6+2/6.7+.....+2/15.16
=1/2[1/4.5+1/5.6+1/6.7+.....+1/15.16]
=1.2[1/4-1/5+1/5-1/6+.....+1/15-1/16]
=1/2[1/4-1/16]
=1/2.3/16
=3/32

Ta có:
\(A=\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+...+\frac{6}{87.90}\)
\(A=2.\left(\frac{3}{15.18}+\frac{3}{18.21}+\frac{3}{21.24}+...+\frac{3}{87.90}\right)\)
\(A=2.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(A=2.\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(A=2.\frac{1}{18}=\frac{1}{9}\)

Có: \(N=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+....+\frac{1}{120}\)
\(=>N=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)
\(=>N=\frac{2}{4\cdot5}+\frac{2}{5\cdot6}+\frac{2}{6\cdot7}+...+\frac{2}{15\cdot16}\)
\(=>N=\left(\frac{2}{4}-\frac{2}{5}+\frac{2}{5}-\frac{2}{6}+...+\frac{2}{15}-\frac{2}{16}\right)\)
\(=>N=\frac{2}{4}-\frac{2}{16}\)
\(=>N=\frac{1}{2}-\frac{1}{8}\)
\(=>N=\frac{8-2}{16}=\frac{6}{16}=\frac{3}{8}\)
Vậy \(N=\frac{3}{8}\)
Ta có :
\(N=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)
\(N=2\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)\)
\(N=2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)
\(N=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(N=2\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(N=\frac{1}{2}-\frac{1}{8}\)
\(N=\frac{3}{8}\)
Vậy \(N=\frac{3}{8}\)
Chúc bạn học tốt ~

\(=\frac{2}{20}+\frac{2}{30}+...+\frac{2}{420}\)
\(=\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{20.21}\)
\(=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{20}-\frac{1}{21}\right)\)
\(=2\left(\frac{1}{4}-\frac{1}{21}\right)\)
\(=2\times\frac{17}{84}\)
\(=\frac{17}{72}\)
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