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{...}

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

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

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

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

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

Vì (x - 1)²⁰¹⁶ ≥ 0 ; (y - 2)²⁰¹⁶ ≥ 0 | x + y + z | ≥ 0 với mọi x

Để (x - 1)²⁰¹⁶ + (y + 2)²⁰¹⁶ + | x + y - z | = 0 khi (x - 1)²⁰¹⁶ = 0 ; (y + 2)²⁰¹⁶ = 0; | x + y - z | = 0

<=> x - 1 = 0 và y + 2 = 0 => x = 1 và y = - 2

Thay x = 1 và y = - 2 vào BT : | x + y - z | = 0 ta được :

| 1 - 2 - z | = 0 <=> 1 - 2 - z = 0 <=> - 1 - z = 0 => z = - 1

Vậy x = 1 ; y = - 2 ; z = - 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

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