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\(A=\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+........+\frac{1}{100.104}\)
\(=\frac{1}{4}.\left(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+.......+\frac{4}{100.104}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+.......+\frac{1}{100}-\frac{1}{104}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{5}-\frac{1}{104}\right)\)
\(=\frac{1}{4}.\frac{99}{520}=\frac{99}{2080}\)
Gọi \(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{22.25}\)
\(\Leftrightarrow\)\(3A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{22.25}\)
\(\Leftrightarrow\)\(3A=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{22}-\frac{1}{25}\)
\(\Leftrightarrow\)\(3A=1-\frac{1}{25}\)
\(\Leftrightarrow\)\(3A=\frac{24}{25}\)
\(\Leftrightarrow\)\(A=\frac{24}{25}:3\)
\(\Leftrightarrow\)\(A=\frac{24}{25}.\frac{1}{3}\)
\(\Leftrightarrow\)\(A=\frac{8}{25}\)
Vậy \(A=\frac{8}{25}\)
Đặt \(C=\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{22.25}\)
\(\Rightarrow3C=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{22.25}\)
\(\Rightarrow3C=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{22}-\frac{1}{25}\)
\(\Rightarrow3C=1-\frac{1}{25}=\frac{24}{25}\)
\(\Rightarrow C=\frac{24}{25}:3=\frac{8}{25}\)
Vậy \(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{22.25}=\frac{8}{24}\)
= \(\frac{11}{10}\cdot\frac{12}{11}\cdot\frac{13}{12}\cdot\frac{14}{13}\cdot\frac{15}{14}\cdot\frac{16}{15}\cdot\frac{17}{16}\)
=11/10 x 12/11 x 13/12 x 14/13 x 15/14 x 16/15 x 17/16
= \(\frac{17}{10}\)
=\(\frac{11}{10}\)x \(\frac{12}{11}\)x .......... x \(\frac{16}{15}\)x\(\frac{17}{16}\)
= \(\frac{11^1x12^1x......x16^1x17}{10x11^1x...x15^1x16^1}\)( những số có số nhỏ ở trên là rút gọn với số khác VD:11 rút gọn cho 11 )
=\(\frac{1x1x......x1x17}{10x1x.......x1x1}\)
=\(\frac{17}{10}\)
= 1,7
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}+1\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}+1\)
\(=1-\frac{1}{1000}+1\)
\(=\frac{1000}{1000}-\frac{1}{1000}+\frac{1000}{1000}\)
\(=\frac{1999}{1000}\)
Tham khảo nhé~
Ta có:A: 1/1.2 +1/2.3 +1/3.4+...+1/18.19+1/19.20
=> A= 1-1/2 +1/2-1/3+1/3-1/4+...+1/18-1/19+1/19-1/20
=>A= 1-1/20=19/20
Lời giải:
$A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+....+\frac{19-18}{18.19}+\frac{20-19}{19.20}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}$
$=1-\frac{1}{20}=\frac{19}{20}$
\(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{19.20}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{5}-\frac{1}{20}\)
\(=\frac{4}{20}-\frac{1}{20}=\frac{3}{20}\)