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{...}

Bài này đễ thôi 

Giải |x-2015|²⁰¹⁶ +|x-2016|²⁰¹⁵=1 

Vì |x-2015|²⁰¹⁶ +|x-2016|²⁰¹⁵=1 nên |x-2015|²⁰¹⁶ hoặc |x-2016|²⁰¹⁵ bằng 1 hoặc 0

Nên

-) Nếu |x-2015|²⁰¹⁶ bằng 0 thì 

x = 2015

-) Nếu |x-2015|²⁰¹⁶ bẳng 1 thì 

x = 2016

-) Nếu |x-2016|²⁰¹⁵ bằng 0 thì 

x =  2016

-) Nếu |x-2016|²⁰¹⁵ bằng 1 thì :

x = 2017

Mà x chỉ có một nên trường hợp là |x-2016|²⁰¹⁵ bằng 0 và |x-2015|²⁰¹⁶ bằng 1 

Nên x = 2016

Vậy x = 2016 

\(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+...+\frac{x-2016}{1}=2016\)

\(\Leftrightarrow\frac{x-1}{2016}-1+\frac{x-2}{2015}-1+\frac{x-3}{2014}-1+...+\frac{x-2016}{1}-1=0\)

\(\Leftrightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}+\frac{x-2017}{2014}+...+\frac{x-2017}{1}=0\)

\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+...+1\right)=0\)

Có: \(\frac{1}{2016}+\frac{1}{2015}+...+1\ne0\)

\(\Rightarrow x-2017=0\)

\(\Rightarrow x=2017\)

<=> \(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+....+\frac{x-2016}{1}-2016=0\)\(=0\)

<=> \(\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2015}-1\right)+...+\left(\frac{x-2016}{1}-1\right)=0\)

<=> \(\frac{x-2017}{2016}+\frac{x-2017}{2015}+...+\frac{x-2017}{1}=0\)

<=> \(\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+...+\frac{1}{1}\right)=0\)

<=> \(x-2017=0\)\(\left(do\frac{1}{2016}+\frac{1}{2015}+...+\frac{1}{1}>0\right)\)

<=> \(x=2017\)

Vậy x = 2017

đúng thì

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

x là số đối của các số trong ngoặc