2:

=1-1+1-1=0

3:

a: =>34*(100+1)/2:a=17

=>a=101

b: =>5/3(x-1/2)=5/4

=>x-1/2=5/4:5/3=3/4

=>x=5/4

1a, \(\dfrac{2005}{2001}\) = 1+\(\dfrac{4}{2001}\); \(\dfrac{2009}{2005}\)=1+\(\dfrac{4}{2005}\)vì\(\dfrac{4}{2001}\)>\(\dfrac{4}{2005}\)nên\(\dfrac{2005}{2001}\)>\(\dfrac{2009}{2005}\)

1b,\(\dfrac{1313}{1515}\)=\(\dfrac{1313:101}{1515:101}\)= \(\dfrac{13}{15}\); \(\dfrac{131313}{151515}\)=\(\dfrac{131313:10101}{151515:10101}\)=\(\dfrac{13}{15}\)

Vậy \(\dfrac{13}{15}\)=\(\dfrac{1313}{1515}\)=\(\dfrac{131313}{151515}\)

\(\dfrac{1}{7}\times\dfrac{21}{8}-\dfrac{3}{8}\times\dfrac{1}{7}-\dfrac{1}{7}\times\dfrac{2}{8}\\ =\dfrac{1}{7}\times\left(\dfrac{21}{8}-\dfrac{3}{8}-\dfrac{2}{8}\right)\\ =\dfrac{1}{7}\times\dfrac{16}{8}\\ =\dfrac{1}{7}\times2\\ =\dfrac{2}{7}\)

\(\dfrac{1}{7}\times\dfrac{21}{8}-\dfrac{3}{8}\times\dfrac{1}{7}-\dfrac{1}{7}\times\dfrac{2}{8}\)

=\(\dfrac{1}{7}\times\left(\dfrac{21}{8}-\dfrac{3}{8}-\dfrac{2}{8}\right)\)

=\(\dfrac{1}{7}\times2\)

=\(\dfrac{2}{7}\)

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\)

\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)

\(=\dfrac{1}{1}\cdot\dfrac{1}{2}+\dfrac{1}{2}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{1}{4}+\dfrac{1}{4}\cdot\dfrac{1}{5}+\dfrac{1}{5}\cdot\dfrac{1}{6}+\dfrac{1}{6}\cdot\dfrac{1}{7}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)

\(=\dfrac{1}{1}-\dfrac{1}{7}=\dfrac{7}{7}-\dfrac{1}{7}=\dfrac{6}{7}\)

Bài 1:

1: \(B=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)

\(=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\cdots+\frac{1}{11\cdot13}\)

\(=\frac12\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\cdots+\frac{2}{11\cdot13}\right)\)

\(=\frac12\left(\frac13-\frac15+\frac15-\frac17+\cdots+\frac{1}{11}-\frac{1}{13}\right)\)

\(=\frac12\left(\frac13-\frac{1}{13}\right)=\frac12\cdot\frac{10}{39}=\frac{5}{39}\)

2: \(C=\frac12+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}\)

\(=\frac24+\frac{2}{28}+\frac{2}{70}+\frac{2}{130}+\frac{2}{208}+\frac{2}{304}\)

\(=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+\frac{2}{10\cdot13}+\frac{2}{13\cdot16}+\frac{2}{16\cdot19}\)

\(=\frac23\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}+\frac{3}{16\cdot19}\right)\)

\(=\frac23\left(1-\frac14+\frac14-\frac17+\frac17-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}\right)\)

\(=\frac23\left(1-\frac{1}{19}\right)=\frac23\cdot\frac{18}{19}=\frac{2\cdot6}{19}=\frac{12}{19}\)

Bài 2:

\(\frac{1}{101}>\frac{1}{300};\frac{1}{102}>\frac{1}{300};\ldots;\frac{1}{299}>\frac{1}{300};\frac{1}{300}=\frac{1}{300}\)

Do đó: \(\frac{1}{101}+\frac{1}{102}+\cdots+\frac{1}{300}>\frac{1}{300}+\frac{1}{300}+\cdots+\frac{1}{300}=\frac{200}{300}=\frac23\) (ĐPCM)

khó thế toy ko bt =)))

toy cũng vậy

\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....+\frac{1}{20}.\left(1+2+....+20\right)\)

\(=1+\frac{1}{2}\times\frac{2.3}{2}+\frac{1}{3}\times\frac{3.4}{2}+...+\frac{1}{20}\times\frac{20.21}{2}\)

\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}\)

\(=\frac{\left(2+21\right).20:2}{2}=\frac{230}{2}=115\)

Số cuối là

\(\frac{1}{10}.\left(1+2+3+...+10\right)\) hay \(\frac{1}{20}.\left(1+2+3+...+20\right)\) ??

a= 1/2+1/2^2+.............1/2^10

2a=1+1/2+1/2^2+............+1/^9

2a-a=1-1/2^10

=1-1/2004

=2003/2004

18?

190:))