\(\left(\left|x\right|+2017\right).\left(504.\left|x\right|-2016\right)< 0\)    (*)

Nhận thấy:  \(\left|x\right|+2017>0\)

Nên từ (*) suy ra:   \(504.\left|x\right|-2016< 0\)

                      <=>   \(504.\left|x\right|< 2016\)

                      <=> \(\left|x\right|< 4\)

Do x là số nguyên nên:  \(x=\left\{0;\pm1;\pm2;\pm3\right\}\)

Đường Quỳnh Giang : mơn nhiuuuuuuuuuuu!!!!!!!!

\(\left(\left|x\right|+2017\right)\left(504\left|x\right|-2016\right)< 0\)

\(\Leftrightarrow\left|x\right|+2017\)và \(504\left|x\right|-2016\)trái dấu

mà \(\left|x\right|+2017>0\forall x\)

\(\Leftrightarrow504\left|x\right|-2016< 0\)

\(\Leftrightarrow504\left|x\right|< 2016\)

\(\Leftrightarrow\left|x\right|< 4\)

\(\Leftrightarrow-4< x< 4\) mà x là số nguyên 

\(\Leftrightarrow x\in\left\{-3;-2;-1;0;1;2;3\right\}\)

Bg

Ta có: (|x| + 2017)(504|x| - 2016) < 0  (x\(\inℤ\))

Mà |x| + 2017 > 0 

Để biểu thức < 0 thì 504|x| - 2016 < 0

=> 504|x| < 2016

=> |x| < 4

=> |x| \(\in\){0; 1; 2; 3}

=> x \(\in\){0; 1; -1; 2; -2; 3; -3}

Vậy x \(\in\){0; 1; -1; 2; -2; 3; -3}

Vì |x| + 2017 \(\ge2017>0\forall x\)

=> 504|x| - 2016 < 0

=> 504|x| < 2016

=> |x| < 4

=> -4 < x < 4

=> \(x\in\left\{-3;-2;-1;0;1;2;3\right\}\)(Vì x nguyên)

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

Vì (2017;2016) =1

=> x >/ 0

và y>/ -1 

thì (2017^x - 2016^y+1 là số nguyên