\(\frac{3}{2}>1;\frac{1212}{3232}< 1\)\(\Rightarrow\frac{3}{2}>\frac{1212}{3232}\)

Ta có:

\(\frac{3}{2}>\frac{2}{2}=1.\)

Mà \(\frac{1212}{3232}< \frac{3232}{3232}=1.\)

=>\(\frac{3}{2}>1>\frac{1212}{3232}\)

Vậy \(\frac{3}{2}>\frac{1212}{3232.}\)

Ko cần tk mk đâu mk tl cho vui thôi nhá.

-Chúc mọi người vui vẻ-

\(\frac{1212}{3232}\)\(=\frac{3}{8}\)

\(\frac{3}{2}>1\)

\(\frac{3}{8}< 1\)

\(\Rightarrow\frac{3}{2}>\frac{3}{8}\)

1 < 1212 < 1414 < 3232

là phân số hay số tự nhiên vậy  bạn?

\(\frac{2323+1313+1414+1515+1616}{4141+2222+2323+2424+2525}\)

=\(\frac{8181}{13635}\)

18180,8               

\(\frac{1010+1111+1212+1313+1414+1515}{2020+2121+2222+2323+2424+2525}=\frac{10\times101+11\times101+12\times101+13\times101+14\times101+15\times101}{20\times101+21\times101+22\times101+23\times101+24\times101+25\times101}=\frac{101\times\left(10+11+12+13+14+15\right)}{101\times\left(20+21+22+23+24+25\right)}\)

\(=\frac{10+11+12+13+14+15}{20+21+22+23+24+25}=\frac{75}{135}=\frac{5}{9}\)

1010+1111+1212+1313+1414+1515= 7575

2020+2121+2222+2323+2424+2525= 13635

có 1 dấu gạch ngang mờ nhạt kìa bạn

khong hieu

\(\frac{1010+1111+1212+1313+1414+1515+1616+1717}{2020+2121+2222+2323+2424+2525+2626+2727}\)

\(=\frac{101.10+101.11+...+101.17}{101.20+101.21+...+101.27}\)

\(=\frac{101.\left(10+11+...+17\right)}{101.\left(20+21+...+27\right)}\)

\(=\frac{108}{188}\)

\(=\frac{27}{47}\)

\(2>\left(\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{4}{96}\right)\cdot5.y>\frac{5}{6}\)

\(\Rightarrow2>\left(\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{1}{24}\right):5.y>\frac{5}{6}\)

\(\Rightarrow2>\left(\frac{20}{120}+\frac{16}{120}+\frac{9}{120}+\frac{5}{120}\right):5.y>\frac{5}{6}\)

\(\Rightarrow2>\frac{5}{12}:5.y>\frac{5}{6}\)

\(\Rightarrow2>\frac{1}{12}.y>\frac{5}{6}\)

Đặt :\(\frac{1}{12}.y=2\Rightarrow y=2:\frac{1}{12}=24\)

\(\frac{1}{12}.y=\frac{5}{6}\Rightarrow y=\frac{5}{6}:\frac{1}{12}=10\)

\(\Rightarrow24>y>10\)

\(\Rightarrow y\in\left\{11;12;...;23\right\}\)

ko biết

I am sorry

Kb nhé

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

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

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

Giải:

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

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

\(=12.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}\right)\) 

\(=12.\dfrac{1}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\right)\) 

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

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

\(=6.\dfrac{8}{33}\) 

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

\(=\dfrac{12}{15}+\dfrac{12}{35}+\dfrac{12}{63}+\dfrac{12}{99}=12\cdot\left[\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}\right]=12\cdot\left[\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+\dfrac{1}{9\cdot11}\right]=12\cdot\dfrac{1}{2}\cdot\left[\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right]=6\cdot\left[\dfrac{1}{3}-\dfrac{1}{11}\right]=6\cdot\dfrac{8}{33}=\dfrac{48}{33}=\dfrac{16}{11}\)

Lời giải:

$\frac{102102}{405405}=\frac{1001\times 102}{1001\times 405}=\frac{102}{405}=\frac{3\times 34}{3\times 135}=\frac{34}{135}$

= 34/125 bạn hãy vote cho mk nhé

Iiiri did we get our first Christmas gift today at 

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

= 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

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

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

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

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

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

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

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

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

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