= \(\frac{12}{15}\) +\(\frac{12}{35}\)+\(\frac{12}{63}\)+\(\frac{12}{99}\)

= 12 x (\(\frac{1}{15}\)+\(\frac{1}{35}\)+\(\frac{1}{63}\)+\(\frac{1}{99}\))

= 12 x ( \(\frac{1}{3x5}\)+\(\frac{1}{5x7}\)+\(\frac{1}{7x9}\)+\(\frac{1}{9x11}\))

= 12 x \(\frac{1}{2}\) x    ( \(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{9}\)+\(\frac{1}{9}\)-\(\frac{1}{11}\))

= 6 x (   \(\frac{1}{3}\) -  \(\frac{1}{11}\))

=  6 x \(\frac{8}{33}\)

= \(\frac{48}{33}\)=\(\frac{16}{11}\)

      Nhớ tk nha

lấy tất cả chia cho 1010

\(\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}\)

\(=1212.\left(\frac{1}{15.101}+\frac{1}{35.101}+\frac{1}{63.101}+\frac{1}{99.101}\right)\)

\(=12.101.\frac{1}{101}.\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)

\(=6.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(=6.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=6.\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=6.\left(\frac{11}{33}-\frac{3}{33}\right)\)

\(=6.\frac{8}{33}\)

\(=\frac{16}{11}\)

Mình giải theo kiểu lớp 6 nhá !

=\(33.\left(\frac{34}{15}+\frac{34}{35}+\frac{34}{63}+\frac{34}{99}\right)\)

=\(33.\left[34.\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\right]\)

=\(33.\left[34.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\right]\)

=\(33.\left[34.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).\frac{1}{2}\right]\)

=\(33.\left[34\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right).\frac{1}{2}\right]\)

=\(33.\left[34.\left(\frac{1}{3}-\frac{1}{11}\right).\frac{1}{2}\right]\)

=\(33.\left(34.\frac{8}{33}.\frac{1}{2}\right)\)

=\(33.\frac{136}{33}\)

=\(\frac{33.136}{33}\)(*)

=\(136\)

(Bạn có thể bỏ bước có dấu *)

 (  3434/3535 + 3434/6363 + 3434/9999 ) × 55

= (  34/35 + 34/63 + 34/99 ) × 55

= 34 × ( 1/35 + 1/63 + 1/99 ) × 55

=  34 × 3/55 × 55

= 102/55 × 55

= 102

Mình ko chắc lắm đâu nhá

a,\(\frac{2015.2016+2015-1}{2014+2015.2016}=\frac{2015.2016+2014}{2014+2015.2016}=1\)\(1\)

b,\(=1-\frac{1}{5}+\frac{1}{5}...-\frac{1}{2011}+\frac{1}{2011}-\frac{1}{2015}=1-\frac{1}{2015}=\frac{2014}{2015}\)

c,\(=\frac{12}{35}+\frac{12}{35}+\frac{12}{35}+\frac{12}{35}=\frac{12}{35}.4=\frac{48}{35}\)

48/35 nha

b>=1-1/5+1/5-1/9+...+1/2011-1/2015

 =1-1/2015

=2014/2015

c>=12/35+12/35+12/35+12/35

=12/35x4

=48/35

C= 101 . 35 . 10101.23 / 35.10101 . 23. 101 = 1

Do D = 3535/3534 > 1 => C<D

\(\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}\)

\(=\frac{12}{15}+\frac{12}{35}+\frac{12}{63}+\frac{12}{99}\)

\(=12\cdot\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)

\(=12\cdot\frac{4}{33}\)

\(=\frac{16}{11}\)

\(\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}\)

\(=\frac{12}{15}+\frac{12}{35}+\frac{12}{63}+\frac{12}{99}\)

\(=12\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)

\(=12\left(\frac{1}{3\cdot5}+\frac{1}{3\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}\right)\)

\(=12\cdot\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{3\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}\right)\)

\(=6\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=6\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=6\cdot\frac{8}{33}\)

\(=\frac{48}{33}\)

Sorry , mình làm hơi tắt 1 xíu

Ta có : \(12\cdot\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)

\(=6\cdot\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\right)\)

\(=6\cdot\left(\frac{2}{3\cdot5}+\cdot\cdot\cdot+\frac{2}{9\cdot11}\right)\)

\(=6\cdot\left(\frac{1}{3}-\frac{1}{5}+\cdot\cdot\cdot+\frac{1}{9}-\frac{1}{11}\right)\)

\(=6\cdot\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=6\cdot\frac{8}{33}\)

\(=\frac{16}{11}\)

= 1x1x1

=1

1616/1515 + 1616/3535 + 1616/6363 + 1616/9999

= 32/21 + 1616/6363 + 1616/9999

= 16/9 + 1616/9999

= 64/33

còn cách khác ko bạn

cách khác nhé:

    \(\frac{1616}{1515}+\frac{1616}{3535}+\frac{1616}{6363}+\frac{1616}{9999}\)

   \(=\frac{16}{15}+\frac{16}{35}+\frac{16}{63}+\frac{16}{99}\)

   \(=16\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)

  \(=16\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\)

  \(=8\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

  \(=8\left(\frac{1}{3}-\frac{1}{11}\right)\)

  \(=\frac{64}{33}\)