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24 tháng 7 2016

\(D=\frac{202202}{1212}+\frac{202202}{2020}+\frac{202202}{3030}+\frac{202202}{4242}+\frac{202202}{5656}\)

\(D=\frac{2002.101}{101.12}+\frac{2002.101}{20.101}+\frac{2002.101}{30.101}+\frac{2002.101}{42.101}+\frac{2002.101}{56.101}\)

\(D=\frac{2002}{12}+\frac{2002}{20}+\frac{2002}{30}+\frac{2002}{42}+\frac{2002}{56}\)

\(D=\frac{1001}{6}+\frac{1001}{10}+\frac{1001}{15}+\frac{143}{3}+\frac{143}{4}\)

\(D=\frac{5005}{12}\)

đáp án là 5005/12 nhé bạn

tích cho mik nha

23 tháng 2 2017

\(=\frac{2002}{12}+\frac{2002}{20}+\frac{2002}{30}+\frac{2002}{42}+\frac{2002}{56}\)

\(=2002.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\right)\)

\(=2002.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)

\(=2002.\left(\frac{1}{3}-\frac{1}{8}\right)\)

\(=2002.\frac{5}{24}\)

\(=\frac{5005}{12}\)

23 tháng 2 2017

5005/12

tk mình

25 tháng 3 2015

A=202202.1/1212+202202.1/2020+202202.1/3030+202202.1/4242+202202.1/5656

A=202202.(1/1212+1/2020+1/3030+1/4242+1/5656)

A=202202.5/2424

A=417/1/12

25 tháng 3 2015

A=202202.1/1212+202202.1/2020+202202.1/3030+202202.1/4242+202202.1/5656

A=202202.(1/1212+1/2020+1/3030+1/4242+1/5656)

A=202202.5/2424

A=5005/12

8 tháng 8 2016

\(A=\frac{1}{6.10}+\frac{1}{10.14}+\frac{1}{14.18}+\frac{1}{18.22}+\frac{1}{22.26}+\frac{1}{26.30}\)

  \(=\frac{1}{4}.\left(\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+\frac{1}{18}-\frac{1}{22}+\frac{1}{22}-\frac{1}{26}+\frac{1}{26}-\frac{1}{30}\right)\)

     \(=\frac{1}{4}.\left(\frac{1}{6}-\frac{1}{30}\right)=\frac{1}{4}.\frac{2}{15}=\frac{1}{30}\)

\(B=\frac{5}{2.3}+\frac{5}{3.4}+\frac{5}{4.5}+...+\frac{5}{8.9}\)\(=5.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}\right)\)     \(=5.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}\right)\)

  \(=5.\left(\frac{1}{2}-\frac{1}{9}\right)=5.\frac{7}{18}=\frac{35}{18}\)

\(C=\left(\frac{7^2}{2.9}+\frac{7^2}{9.16}+....+\frac{7^2}{65.72}\right):\left(\frac{1}{3}-\frac{7}{36}\right)\)

   \(=7.\left(\frac{7}{2.9}+\frac{7}{9.16}+...+\frac{7}{65.72}\right):\frac{5}{36}\) \(=7.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{65}-\frac{1}{72}\right):\frac{5}{36}\)'

    \(=7.\left(\frac{1}{2}-\frac{1}{72}\right):\frac{5}{36}=7.\frac{35}{72}:\frac{5}{36}=\frac{49}{2}\)

\(D=\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}+\frac{2}{38.39.40}\)

     \(=2.\left(\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}+\frac{1}{38.39.40}\right)\)

     \(=2.\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}+\frac{1}{38.39}-\frac{1}{39.40}\right)\)

        \(=\frac{1}{2.3}-\frac{1}{39.40}=\frac{259}{1560}\)

\(E=\frac{202202}{1212}+\frac{202202}{2020}+\frac{202202}{3030}+\frac{202202}{4242}+\frac{202202}{5656}\)

    \(=202202.\left(\frac{1}{3.4.101}+\frac{1}{4.5.101}+\frac{1}{5.6.101}+\frac{1}{6.7.101}+\frac{1}{7.8.101}\right)\)

      \(=2002.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\right)\)

        \(=2002.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)

         \(=2002.\left(\frac{1}{3}-\frac{1}{8}\right)=2002.\frac{5}{24}=\frac{5005}{12}\)

     

    

7 tháng 9 2017

1212 / 1313 = 12 / 13

201201 / 202202 = 6097 / 6094

7 tháng 9 2017

ko biết

bạn có phân số nào bé hơn ko gửi đi mik làm co ahihi

14 tháng 7 2018

a,3^200 và 2^300

3^200=(3^2)^100=9^100

2^300=(2^3)^100=8^100

Vì 9^100>8^100=>3^200>2^300

Vậy 3^200>2^300

b, 71^50 và 37^75

71^50=(71^2)^25=5041^25

37^75=(37^3)^25=50653^25

Vì 5041^25<50653^25=> 71^50<37^75

Vậy  71^50<37^75

c, 201201/202202 và 201201201/202202202

201201201/202202202=201201/202202

=> 201201/202202=201201201/202202202

Vậy 201201/202202=201201201/202202202

14 tháng 7 2018

a)

Ta có:3200=32.100=(32)100=9100

2300=23.100=(23)100=8100

Vì 9100>8100

Nên 3200>2300

b) 

Ta có: 7150=712.25=(712)25=504125

3775=373.25=(373)25=5065325

Vì 504125<5065325

Nên 7150<3775

c)

Ta có:

201201/202202=201.1001/202.1001=201/202

201201201/202202202=201.1001001/202.1001001001= 201/202

Vì 201/202=201/202

Nên 201201/202202=201201201/202202202

10 tháng 7 2017

\(M=\frac{7}{4}\times\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}+\frac{3333}{5656}\right)\)

\(M=\frac{7}{4}\times\left(\frac{11}{4}+\frac{33}{20}+\frac{11}{10}+\frac{11}{14}+\frac{33}{56}\right)\)

\(M=\frac{7}{4}\times\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}+\frac{33}{56}\right)\)

\(M=\frac{7}{4}\times\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}+\frac{33}{7.8}\right)\)

\(M=\frac{7}{4}\times\left[33\cdot\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\right]\)

\(M=\frac{7}{4}\times\left[33\times\left(\frac{1}{3}-\frac{1}{8}\right)\right]\)

\(M=\frac{7}{4}\times\left(33\times\frac{5}{24}\right)=\frac{7}{4}\times\frac{55}{8}=\frac{385}{32}\)

10 tháng 3 2016

a. 3200 = (32)100 = 9100

2300 = (23)100 = 8100

Vì 9100 > 8100 => 3200 > 2300

20 tháng 5 2022

ét ô ét

 

20 tháng 5 2022

Ta có:\(\dfrac{201201}{202202}\)=\(\dfrac{201}{202}\);\(\dfrac{201201201}{202202202}\)=\(\dfrac{201}{202}\)

=>\(\dfrac{201}{202}\)=\(\dfrac{201}{202}\) 

=> \(\dfrac{201201}{202202}\)=\(\dfrac{201201201}{202202202}\)

Vậy: \(\dfrac{201201}{202202}\)=\(\dfrac{201201201}{202202202}\)

                                        (quá dễ)

14 tháng 9 2017

\(\frac{201201}{202202}=\frac{201}{202}\)

14 tháng 9 2017

=201/202