13/50 + 9% + 41/100 + 0,24

= 0,26 + 0,09 + 0,41 + 0,24

= (0,26 + 0,24) + (0,09 + 0,41)

= 0,5 + 0,5

= 1

Chúc bạn học tốt.

😁😁😁

= 90091/100 hoặc 900,91

\(=\frac{-\frac{1}{9}+1-\frac{2}{10}+1-\frac{3}{11}+1-...-\frac{92}{100}+1}{\frac{1}{9}+\frac{1}{10}+...+\frac{1}{100}}\)

\(=\frac{\frac{8}{9}+\frac{8}{10}+\frac{8}{11}+...+\frac{8}{100}}{\frac{1}{9}+\frac{1}{10}+...+\frac{1}{100}}\)

\(=\frac{8\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)}{\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}}\)

= 8

\(\left(\frac{5}{13}-\frac{9}{14}\right)-\left(2+-\frac{21}{13}-\frac{5}{-14}\right)\)

\(=-\frac{47}{182}-\frac{135}{182}\)

\(=-1\)

tíc mình nha

(5/13 −9/14 )−(2+−21/13 −5/−14 )

=−47/182 −135/182 

=−1

1/3.(4/13-9/11)+1/3.(9/13-4/22)

=1/3(4/13-9/11+9/13-2/11)

=1/3(4/13+9/13-9/11-2/11)

=1/3(1-1)

=1/3.0=0

Ta thấy : \(\frac{1}{11}>\frac{1}{100},\frac{1}{12}>\frac{1}{100},...,\frac{1}{100}=\frac{1}{100}\)

\(\Rightarrow\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{90}{100}=\frac{9}{10}\)

\(\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{100}>\frac{9}{10}+\frac{1}{10}=1\)

Do đó : \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{100}>1\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{7}+...+\frac{1}{32}-\frac{1}{81}\)

\(=1-\frac{1}{81}\)

\(=\frac{80}{81}\)

a, 1

b,14,1

c,7,404

\(C=\frac{2}{4.7}-\frac{3}{5.9}+\frac{2}{7.10}-\frac{3}{9.13}+...+\frac{2}{301.304}-\frac{3}{401.405}\)

\(C=\left(\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{301.304}\right)-\left(\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{401.405}\right)\)

\(C=\frac{2}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{301.304}\right)-\frac{3}{4}\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{401.405}\right)\)

\(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{301}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+..+\frac{1}{401}-\frac{1}{405}\right)\) \(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{405}\right)\)

\(C=\frac{25}{152}-\frac{4}{27}\)

\(C=\frac{67}{4104}\)

ra 67/1014