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{...}

\(\dfrac{1-2x}{2017}+\dfrac{2-2x}{2016}=\dfrac{3-2x}{2015}+\dfrac{4-2x}{2014}\)

\(\Rightarrow\left(\dfrac{1-2x}{2017}+1\right)+\left(\dfrac{2-2x}{2016}+1\right)=\left(\dfrac{3-2x}{2015}+1\right)+\left(\dfrac{4-2x}{2014}+1\right)\)

\(\Rightarrow\dfrac{2018-2x}{2017}+\dfrac{2018-2x}{2016}-\dfrac{2018-2x}{2015}-\dfrac{2018-2x}{2014}=0\)

\(\Rightarrow\left(2018-2x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)

Vì \(2017>2016>2015>2014\) nên

\(\dfrac{1}{2017}< \dfrac{1}{2016}< \dfrac{1}{2015}< \dfrac{1}{2014}\)

\(\Rightarrow\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}< 0\)

\(\Rightarrow2018-2x=0\Rightarrow x=1009\)

Vậy...........

Chúc bạn học tốt!!!

\(\dfrac{1-2x}{2017}+\dfrac{2-2x}{2016}=\dfrac{3-2x}{2015}+\dfrac{4-2x}{2014}\)

\(\Rightarrow\left(\dfrac{1-2x}{2017}+1\right)+\left(\dfrac{2-2x}{2016}+1\right)=\left(\dfrac{3-2x}{2015}+1\right)+\left(\dfrac{4-2x}{2014}+1\right)\)

\(\Rightarrow\dfrac{2018-2x}{2017}+\dfrac{2018-2x}{2016}-\dfrac{2018-2x}{2015}-\dfrac{2018-2x}{2014}=0\)

\(\Rightarrow\left(20418-2x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)

\(Ta\) \(có\)\(:\) \(2017>2016>2015>2014\)

\(\Rightarrow\dfrac{1}{2017}< \dfrac{1}{2016}< \dfrac{1}{2015}< \dfrac{1}{2014}\)

\(\Rightarrow\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}< 0\)

\(\Rightarrow2018-2x=0\)

\(\Rightarrow2x=2018-0\)

\(\Rightarrow2x=2018\)

\(\Rightarrow x=2018:2\)

\(\Rightarrow x=1009\)

a) Ta có: \(12=2^2.3=4.3=3.4\)

Vậy \(x=3\)

b) hình như đề thiếu

ko có thiếu