Mình giải theo lớp 6

  \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}\)

 \(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{2016}-\frac{1}{2017}\)

  Ta loại các cặp số đổi của nhau như : \(-\frac{1}{2}\)và \(\frac{1}{2}\)thì còn

\(\frac{1}{1}-\frac{1}{2017}\)

\(=\frac{2016}{2017}\)

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.......+1/2016-1/2017

=1-1/2017

=2016/2017

xong rồi bạn ạ

=1/1-1/2+1/2-1/3+...+1/2016-1/2017

= 1/1-1/2017

=2016/2017

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B = \(\frac{3^2}{2.4}+\frac{3^2}{4.6}+\frac{3^2}{6.8}+...+\frac{3^2}{198.200}\)

B = \(\frac{3^2}{2}.\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{3^2}{2}.\left(\frac{1}{4}-\frac{1}{6}\right)+\frac{3^2}{2}.\left(\frac{1}{6}-\frac{1}{8}\right)+...+\frac{3^2}{2}.\left(\frac{1}{198}-\frac{1}{200}\right)\)

B = \(\frac{3^2}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{198}-\frac{1}{200}\right)\)

B = \(\frac{9}{2}.\left(\frac{1}{2}-\frac{1}{200}\right)\)

B = \(\frac{9}{2}.\frac{99}{200}\)

B = \(\frac{891}{400}\)

D = 1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + ... + 48 x 49

3D = 1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3 + 4 x 5 x 3 + ... + 48 x 49 x 3

3D = 1 x 2 x 3 + 2 x 3 x ( 4 - 1 ) + 3 x 4 x ( 5 - 2 ) + 4 x 5 x ( 6 - 3 ) + ... + 48 x 49 x ( 50 - 47 )

3D = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + 4 x 5 x 6 - 3 x 4 x 5 + ... + 48 x 49 x 50 - 47 x 48 x 49

3D = 48 x 49 x 50

D = ( 48 x 49 x 50 ) : 3

D = 39200

E = 1² + 2² + 3² + ... + 48²

E = 1 x 1 + 2 x 2 + 3 x 3 + ... + 48 x 48

E = 1 x ( 2 - 1 ) + 2 x ( 3 - 1 ) + 3 x ( 4 - 1 ) + ... + 48 x ( 49 - 1 )

E = 1 x 2 - 1 + 2 x 3 - 2 + 3 x 4 - 3 + ... + 48 x 49 - 49

E = ( 1 x 2 + 2 x 3 + 3 x 4 + ... + 48 x 49 ) - ( 1 + 2 + 3 + ... + 49 )

Ta tính được vế trong ngoặc thứ nhất là 39200 , còn vế trong ngoặc thứ hai là 1225

thay vào ta được :

E = 39200 - 1225

E = 37975 

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

\(\Rightarrow A=1-\frac{1}{2^{100}}\)

giúp mình với

b) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)

\(=\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{2015-2013}{2013.2015}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2015}\right)=\frac{1007}{2015}\)

Phương trình tương đương với: 

\(\frac{1007X}{2015}=\frac{4}{2015}\Leftrightarrow X=\frac{4}{1007}\)

c) \(\frac{x+1}{2015}+\frac{x+2}{2016}=\frac{x+3}{2017}+\frac{x+4}{2018}\)

\(\Leftrightarrow\frac{x+1}{2015}-1+\frac{x+2}{2016}-1=\frac{x+3}{2017}-1+\frac{x+4}{2018}-1\)

\(\Leftrightarrow\frac{x-2014}{2015}+\frac{x-2014}{2016}=\frac{x-2014}{2017}+\frac{x-2014}{2018}\)

\(\Leftrightarrow x-2014=0\)

\(\Leftrightarrow x=2014\)

a, Bn quy đồng rồi làm nha

b,Có  A=2017^2017+1/2017^2018+1

--> 2017A=2017^2018+2017/2017^2018+1

2017A=2017^2018+1/2017^2018+1 + 2016/2017^2018+1

2017A=1+ 2016/2017^2018+1

Có B=2017^2016+1/2017^2017+1

--> 2017B=2017^2017+2017/2017^2017+1

2017B=2017^2017+1/2017^2017+1 + 2016/2017^2017+1

2017B=1+2016/2017^2017+1

        Vì 1+2016/2017^2018+1 < 1+2016/2017^2017+1

nên 2017A<2017B

-->A<B

A.Ta có : 

\(A=-\frac{15}{46}>-\frac{15}{45}=-\frac{51}{153}>-\frac{51}{151}=B\)

\(\Rightarrow A>B\)

Câu b, giống bạn Kelly Gaming TV