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a) \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{5}-\frac{1}{10}\)
\(=\frac{1}{10}\)
b) \(\frac{2}{10.12}+\frac{2}{12.14}+\frac{2}{14.16}+...+\frac{2}{998.1000}\)
\(=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}-\frac{1}{16}+...+\frac{1}{998}-\frac{1}{1000}\)
\(=\frac{1}{10}-\frac{1}{1000}\)
\(=\frac{99}{1000}\)
c) \(\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{69.90}\)
\(=4.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{89.90}\right)\)
\(=4.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{89}-\frac{1}{90}\right)\)
\(=4.\left(1-\frac{1}{90}\right)\)
\(=4.\frac{89}{90}\)
\(=\frac{178}{45}\)
_Chúc bạn học tốt_
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{15.16}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{15}-\frac{1}{16}\)
\(=1-\frac{1}{16}=\frac{15}{16}\)
\(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{15x16}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{15}-\frac{1}{16}\)
\(=1-\frac{1}{16}\)
\(=\frac{15}{16}\)
Đáp án là 0 bạn nhá !
Thử quy đồng số 2/5 ra mẫu số là 10 rồi trừ y như trừ số tự nhiên đó !
Vậy thôi hà !
S=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{2009}\)-\(\frac{1}{2010}\)
S=1-\(\frac{1}{2010}\)
S=\(\frac{2009}{2010}\)
k nha bn
\(S=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2008\times2009}+\frac{1}{2009\times2010}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2009}+\frac{1}{2009}-\frac{1}{2010}\)
\(=1-\frac{1}{2010}\)
\(=\frac{2009}{2010}\)
Vậy \(S=\frac{2009}{2010}\)
Học tốt #
\(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{999x1000}+1\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}+1\)
\(=2-\frac{1}{1000}=\frac{1999}{1000}\)
có thể đây là bài lớp 4 nhưng mình nghĩ là các bạn lớp 5 cũng sẽ khó khăn đó
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{y\times\left(y+1\right)}=\frac{996}{997}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{y}-\frac{1}{y+1}=\frac{996}{997}\)
\(\Leftrightarrow1-\frac{1}{y+1}=\frac{996}{997}\)
\(\Leftrightarrow\frac{1}{y+1}=1-\frac{996}{997}=\frac{1}{997}\)
\(\Leftrightarrow y+1=997\Leftrightarrow y=996\)
Vậy y = 996
\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{15}+2\left(1-\frac{1}{10}\right)\)
\(=\frac{4}{15}+\frac{9}{5}\)
\(=\frac{31}{15}\)
Bài làm :
Ta có :
\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{13\times15}+\frac{2}{1\times2}+\frac{2}{2\times3}+...+\frac{2}{9\times10}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{15}+2\left(1-\frac{1}{10}\right)\)
\(=\frac{31}{15}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(=1-\frac{1}{2011}\)
\(=\frac{2010}{2011}\)
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2010-1/2011
= 1 - 1/2011
= 2010/ 2011
Đáp số: 2010/2011
Chúy ý công thức: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)