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) 1/1.2 + 1/2.3 + 1/3.4 + .... + 1/x.(x+1) = 499/500

1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + .... + 1/x - 1/x+1 = 499/500

1 - 1/x+1 = 499/500

1/x+1 = 1 - 499/500

1/x+1 = 1/500

x + 1 = 500

     x = 500 - 1 

     x = 499

b) 1/1.3 + 1/3.5 + 1/5.7 + .... + 1/x.(x+2) = 20/41

1/2 . [ 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/x.(x+2) ] = 20/41

1/2 . [ 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/x - 1/x+2 ] = 20/41

1/2 . [ 1 - 1/x+2 ) = 20/41

1 - 1/x+2 = 20/41 : 1/2

1 - 1/x+2 = 40/41

1/x+2 = 1 - 40/41

1/x+2 = 1/41

x + 2 = 41

      x = 41 - 2 

      x = 39

Vế trái bằng vế phải nên đẳng thức được chứng minh.

đặt x²⁰¹¹-x²⁰¹⁰+1=t thì đa thức trở thành:

t(t+1)-20=t²+t-20=(t²-5t)+(4t-20)=t(t-5)+4(t-5)=(t-5)(t+4)(*)

thay t= x²⁰¹¹-x²⁰¹⁰+1 vào (*) ta có:

( x²⁰¹¹-x²⁰¹⁰+1-5)( x²⁰¹¹-x²⁰¹⁰+1+4)=( x²⁰¹¹-x^2010-4)( x²⁰¹¹-x²⁰¹⁰+5)

=>( x²⁰¹¹-x²⁰¹⁰+1)( x²⁰¹¹-x²⁰¹⁰+2)-20=( x²⁰¹¹-x^2010-4)( x²⁰¹¹-x²⁰¹⁰+5)

\(A=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)\cdot...\cdot\left(1+\frac{1}{99\cdot101}\right)\)

\(A=\frac{4}{1\cdot3}\cdot\frac{9}{2\cdot4}\cdot\frac{16}{3\cdot5}\cdot...\cdot\frac{10000}{99\cdot101}\)

\(A=\frac{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)\cdot...\cdot\left(100\cdot100\right)}{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)\cdot...\cdot\left(99\cdot101\right)}\)

\(A=\frac{\left(2\cdot3\cdot4\cdot...\cdot100\right)\left(2\cdot3\cdot4\cdot...\cdot100\right)}{\left(1\cdot2\cdot3\cdot...\cdot99\right)\left(3\cdot4\cdot5\cdot...\cdot101\right)}\)

\(A=\frac{100\cdot2}{1\cdot101}\)

\(A=\frac{200}{101}\)

A = 2011 x (2009  + 1) - 1/ 2011 x 2009 + 2010

A = 2011 x 2009 + 2011 x 1 - 1/2011 x 2009 + 2010

A = 2011 x 2009 + 2010/2011 x 2009 + 2010

A = 1

B.