=>2/6+2/12+2/20+...+2/x(x+1)=2013/2015

=>2(1/2.3+1/3.4+1/4.5+...+1/x(x+1)=2013/2015

=>2(1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1)=2013/2015

=>(1/2-1/x+1)=2013/2015:2

=>-(1/x+1)=2013/4030-1/2

=>-(1/x+1)=-(1/2015)=>x+1=2015=>x=2014

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\cdot\left(x+1\right)}=\frac{2013}{2015}\)

\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{x\cdot\left(x+1\right)}\right)=\frac{2013}{2015}\)

\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\cdot\left(x+1\right)}\right)=\frac{2013}{2015}\)

\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2013}{2015}\)

\(\Rightarrow\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2013}{2015}:2\)

\(\Rightarrow-\frac{1}{x+1}=\frac{2013}{4030}-\frac{1}{2}\)

\(\Rightarrow-\frac{1}{x+1}=-\frac{1}{2015}\Rightarrow x+1=2015\Rightarrow x=2014\)

bạn trên trả lời đúng rùi đó

Trần Thị Loan chuẩn rồi 

= 2/(2.3) + 2/3.4 + 2/4.5 +...+ 2/x(x+1)

= 2 [1/2-1/3+1/3-1/4+...+1/x-1/(x+1)]

=2[1/2-1/(x+1)]= (x-1)/(x+1)

= 2001/2003

==> x=2002

x=2002

Mình Giúp Họ Giải Toán Đầu tiên Mà Họ Lại Làm Ngơ sai bét

1/2+1/6+1/12+1/20+...+1/x(x+1)=2015/2016

       1/1.2+1/2.3+1/3.4+.....+1/x.(x+1)=2015/2016

       1-1/2+1/2-1/3+1/3-1/4+......+1/x-1/x+1=2015/2016

         1-1/x-1=2015/2016

            1/x+1=1-2015/2016

               1/x+1=1/2016

 =>               x+1=2016

                 x=2016-1

                  x=2015

vậy x =2015

tích mình nha

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.......+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)

=>\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.......+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)

=>\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+......+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)

=>\(1-\frac{1}{x+1}=\frac{2015}{2016}\)

=>\(\frac{1}{x+1}=1-\frac{2015}{2016}=\frac{1}{2016}\)

=>x+1=2016

=>x=2015

Vậy x=2015

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)

\(1-\frac{1}{x+1}=\frac{2015}{2016}\)

\(\frac{1}{x+1}=1-\frac{2015}{2016}\)

\(\frac{1}{x+1}=\frac{1}{2016}\)

=>x+1=2016

=>x=2015

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{2015}{2016}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2015}{2016}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}:2\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{4032}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4032}=\frac{1}{4032}\)

=> x+1=4032

=> x = 4032 - 1

=> x = 4031

TÍnh S=3/1.4+3/4.7+3?7>!0+...+3/n(n+3) với n là số tự nhiên . chứng minh S<1

câu 1. tìm x nguyên để \(\frac{-35}{6}\)<x<\(\frac{-18}{5}\)

<=> -4,375<x<-3,6

mà x\(\in\)Z nên x={-4}

câu 2. A=\(\frac{2015}{2016}\)+\(\frac{2016}{2017}\)

B=\(\frac{2015+2016}{2016+2017}\)=\(\frac{2015}{2016+2017}\)+\(\frac{2016}{2016+2017}\)

Vì \(\frac{2015}{2016+2017}\)<\(\frac{2015}{2016}\); \(\frac{2016}{2016+2017}\)<\(\frac{2016}{2017}\)

Vậy B<A

cau3:

\(\frac{1}{3}\)+\(\frac{1}{6}\)+\(\frac{1}{10}\)+.....+\(\frac{2}{x\left(x+1\right)}\)=\(\frac{2007}{2009}\)

2.(\(\frac{1}{6}\)+\(\frac{1}{12}\)+\(\frac{1}{20}\)+.....+\(\frac{1}{x\left(x+1\right)}\))=\(\frac{2007}{2009}\)

2.(\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+.....+\(\frac{1}{x\left(x+1\right)}\))=\(\frac{2007}{2009}\)

2.(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{5}\)+.....+\(\frac{1}{x}\)-\(\frac{1}{x+1}\))=\(\frac{2007}{2009}\)

2.(\(\frac{1}{2}\)-\(\frac{1}{x+1}\))=\(\frac{2007}{2009}\)

\(\frac{1}{2}\)-\(\frac{1}{x+1}\)=\(\frac{2007}{4018}\)

\(\frac{1}{x+1}\)=\(\frac{1}{2}\)-\(\frac{2007}{4018}\)

\(\frac{1}{x+1}\)=\(\frac{1}{2009}\)

x+1=2009

x=2009-1

x=2008