thiếu đề =="

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

1-1/2 + 1/2 - 1/3 + ... + 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 = 2015

vậy x = 2015

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

a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\) nên x+1=0

=>x=0-1

=>x-1

a:x+1/10+x+1/11+x+1/12=x+1/13+x+1/14

 <=>(x+1)(1/10 + 1/11+1/12) =(x+1)(1/13 + 1/14) 
<=>(x+1)(1/10 + 1/11+1/12 -1/13 -1/14)=0 
<=> x+1=0(vì biểu thức 1/10 + 1/11 +1/12-1/13-1/14#0) 
<=>x= -1

b:hình như sai đề

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

tích mình nha

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

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

1,-2015

2,50

3,-2015

thiện xạ 5a3 có thể giải chi tiết ra đc k? Mk cần cách lm

2) 1+2+3+...+x=1275

  Có SSH là: (x+1):1+1=x(SH)

  => (x+1).x:2=1275

=>(x+1).x=1275.2

=>(x+1).x=2550

=>(x+1).x=51.50

=>x=50

3) |x+2015|+|x+2016|=1

Ta thấy  |x+2015| và  |x+2016| > hoặc = 0 với mọi x

=> 1= 0+1=1+0

+) x+2015=0=>x=-2015

     x+2016=1=>x=-2015

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

     x+2016=0=> x=-2016

Vậy xE{...}

1/1x2+1/2x3+...+1/x(x+1)=2015/2016

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

2/1-1/x+1=2015/2016

2016/2016-1/x+1=2015/2016

1/x+1=2016/2016-2015/2016

1/x+1=1/2016

x+1=2016

x=2016-1

x=2015

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

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

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

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

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

\(\frac{1}{1x2}\)+\(\frac{1}{2x3}\)+\(\frac{1}{3x4}\)+.....+\(\frac{1}{x\left(x+1\right)}\)=\(\frac{2015}{2016}\)

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

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

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

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

\(\Rightarrow\)x + 1= 2016

\(\Rightarrow\)x = 2015