297/6>306/25                

297/16 > 306 / 25 

cách trinh bày các bạn ơi

\(1,\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}=\frac{12}{15}+\frac{12}{35}+\frac{12}{63}+\frac{12}{99}=6\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\right)=6\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).Tacocongthuc:\frac{1}{n}-\frac{1}{n+k}=\frac{k}{n\left(n+k\right)}\Rightarrow\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}=6\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-.....-\frac{1}{11}\right)=6\left(\frac{1}{3}-\frac{1}{11}\right)=\frac{48}{33}=\frac{16}{11}\)

\(2,\left(x+1\right)+\left(x+2\right)+.....+\left(x+211\right)=211x+\left(1+2+....+211\right)=211x+\frac{212.211}{2}=211x+22366=23632\Leftrightarrow211x=23632-22366=1266\Leftrightarrow x=6\)

a, \(14:\left(4\frac{2}{3}:1\frac{5}{9}\right)+14:\left(\frac{2}{3}+\frac{8}{9}\right)\)

=> \(14:\frac{28}{9}+14:\frac{14}{9}=>14.\frac{9}{28}+14.\frac{9}{14}\)

=> 14. ( \(\frac{9}{28}+\frac{9}{14}\) )

=> \(14.\frac{27}{28}=\frac{419}{28}\)

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

=> \(\frac{4}{5}+\frac{12}{35}+\frac{4}{21}+\frac{4}{33}\)

=> \(\frac{8}{7}+\frac{24}{77}=\frac{16}{11}\)

bài 2 :

( x + 1 ) + ( x + 2 ) + ... + ( x + 211 ) = 23632

=> ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 211 ) = 23632

=> 211x + 22366 = 23632

=> 211x = 23632 - 22366

=> 211x = 1266

=> x = 1266 : 211

x = 6

1 < 1212 < 1414 < 3232

là phân số hay số tự nhiên vậy  bạn?

Ta có:

\(\frac{2017^{10}+1}{2017^{10}-1}=1+\frac{2}{2017^{10}-1}\)

Lại có: 

\(\frac{2017^{10}-1}{2017^{10}-3}=1+\frac{2}{2017^{10}-3}\)

Vì \(1+\frac{2}{2017^{10}-1}< 1+\frac{2}{2017^{10}-3}\)

Nên \(\frac{2017^{10}+1}{2017^{10}-1}< \frac{2017^{10}-1}{2017^{10}-3}\)

Vậy \(\frac{2017^{10}+1}{2017^{10}-1}< \frac{2017^{10}-1}{2017^{10}-3}\)

Ta có

\(\frac{2017^{10}+1}{2017^{10}-1}=\frac{2017^{10}-1+2}{2017^{10}-1}=1+\frac{2}{2017^{10}-1}\)
\(\frac{2017^{10}-1}{2017^{10}-3}=\frac{2017^{10}-3+2}{2017^{10}-3}=1+\frac{2}{2017^{10}-3}\)
\(\Rightarrow1+\frac{2}{2017^{10}-1}< 1+\frac{2}{2017^{10}-1}\)
\(\Rightarrow\frac{2017^{10}+1}{2017^{10}-1}< \frac{2017^{10}-1}{2017^{10}-3}\)

Ta có : \(A=\frac{20^{10}+1}{20^{10}-1}=\frac{\left(20^{10}-1\right)+2}{20^{10}-1}=\frac{20^{10}-1}{20^{10}-1}+\frac{2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)

          \(B=\frac{20^{10}-1}{20^{10}-3}=\frac{\left(20^{10}-3\right)+2}{20^{10}-3}=\frac{20^{10}-3}{20^{10}-3}+\frac{2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)

Do \(20^{10}-1>20^{10}-3\)

\(\Rightarrow\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\Rightarrow1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}\)

Hay : \(A< B\) 

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

= 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Sorry , mình làm hơi tắt 1 xíu

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

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

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

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

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

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

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

bạn kiểm tra lại B xem sao

A=\(\frac{8^{18^{ }}+1}{8^{19}+1}\)=> 8A=8.

Xin lỗi nha mình chưa làm xong