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a) \(\frac{1}{3}-\frac{3}{5}+\frac{5}{7}-\frac{7}{9}+\frac{9}{11}-\frac{11}{13}+\frac{13}{5}+\frac{11}{13}-\frac{9}{11}+\frac{7}{9}-\frac{5}{7}+\frac{3}{5}-\frac{1}{3}\)
=\(\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{5}{7}-\frac{5}{7}\right)+\left(\frac{7}{9}-\frac{7}{9}\right)+\left(\frac{11}{13}-\frac{11}{13}\right)+\frac{13}{5}\)
=\(\frac{0}{3}+\frac{0}{7}+\frac{0}{9}+\frac{0}{13}+\frac{13}{15}\)
=\(0+0+0+0+\frac{13}{5}\)
=\(\frac{13}{5}\)
\(\frac{1}{3}-\frac{3}{5}+\frac{5}{7}-\frac{7}{9}+\frac{9}{11}-\frac{11}{13}+\frac{13}{5}+\frac{11}{13}-\frac{9}{11}+\frac{7}{9}-\frac{5}{7}+\frac{3}{5}-\frac{1}{3}\)
\(=\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{-3}{5}+\frac{3}{5}\right)+\left(\frac{5}{7}-\frac{5}{7}\right)+\left(\frac{-7}{9}+\frac{7}{9}\right)+\left(\frac{-11}{13}+\frac{11}{13}\right)+\frac{13}{5}\)
\(=0+0+0+0+0+\frac{13}{5}\)
\(=\frac{13}{5}\)
Chúc bạn học tốt !
C = 1/100 - ( 1/2.1 + 1/3.2 + ... + 1/98.97 + 1/99.98 + 1/100.99
C = 1/100 - ( 1- 1/2+ 1/2 - 1/3 + ... + 1/97 - 1/98 + 1/98 - 1/99 + 1/99 - 1/100 )
C = 1/100 - ( 1 - 1/100 )
C = 1/100 - 99/100
C = \(\frac{-49}{50}\)
\(=\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{3}{7}-\frac{3}{7}\right)+\left(\frac{5}{9}-\frac{5}{9}\right)+\left(\frac{2}{11}-\frac{2}{11}\right)+\left(\frac{7}{13}-\frac{7}{13}\right)+\frac{9}{16}\)
= 0 + \(\frac{9}{16}\)
= \(\frac{9}{16}\)
tick nha bn
\(B=\frac{1}{5}-\frac{3}{7}+\frac{5}{9}-\frac{2}{11}+\frac{7}{13}-\frac{9}{16}-\frac{7}{13}+\frac{2}{11}-\frac{5}{9}+\frac{3}{7}-\frac{1}{5}\)
\(=\left(\frac{1}{5}-\frac{1}{5}\right)+\left(-\frac{3}{7}+\frac{3}{7}\right)+\left(\frac{5}{9}-\frac{5}{9}\right)+\left(-\frac{2}{11}+\frac{2}{11}\right)+\left(\frac{7}{13}-\frac{7}{13}\right)-\frac{9}{16}\)
\(=0+0+0+0+0-\frac{9}{16}\)
\(=-\frac{9}{16}\)
^...^ 
^_^
Bài 1:
Ta có:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Mà \(\frac{99}{100}< 1\)
\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)