Ta có :

\(\frac{-144}{12}\)\(=-12\)

\(\frac{-40}{5}\)\(=-8\)

\(\Rightarrow-12\le x\le-8\)

x thuộc { -12 ; -11 ; -10 ; -9 ; -8 }

Vậy A = { -12 ; -11 ; -10 ; -9 ; -8 }

A={-12;-11;-10;-9;-8}

Ta có : 

\(\frac{-144}{12}=-12\)

\(\frac{-40}{5}=-8\)

\(\Rightarrow-12\le x\le-8\)

\(\Rightarrow x\in\) { -12 ; -11 ; -10 ; -9 ; -8 }

-2\(\le\)\(\frac{x}{2}\)\(\le\)1

\(\Rightarrow\)-4\(\le\)x\(\le\)2

\(\Rightarrow\)x\(\in\){-4;-3;-2;-1;0;1;2}

Tổng các số nguyên x là:

        [-4+(-3)+(-2)+(-1)+0+1+2]=-7

         Vậy....

Với mọi số thực ta luôn có:

`(x-y)^2>=0`

`<=>x^2-2xy+y^2>=0`

`<=>x^2+y^2>=2xy`

`<=>(x+y)^2>=4xy`

`<=>(x+y)^2>=16`

`<=>x+y>=4(đpcm)`

\(\dfrac{1}{x+3}+\dfrac{1}{y+3}=\dfrac{x+3+y+3}{\left(x+3\right)\left(y+3\right)}\)

\(=\dfrac{x+y+6}{3x+3y+13}\)(vì \(xy=4\))

=> \(\dfrac{x+y+6}{3x+3y+13}\)≤\(\dfrac{2}{5}\)

<=> \(5\left(x+y+6\right)\)≤\(2\left(3x+3y+13\right)\)

<=>\(6x+6y+26-5x-5y-30\)≥\(0\)

<=> \(x+y-4\)≥\(0\)

Áp dụng BĐT AM-GM \(\dfrac{a+b}{2}\)≥\(\sqrt{ab}\)

Ta có \(\dfrac{x+y}{2}\)≥\(\sqrt{xy}\)

<=>\(x+y\) ≥ 2\(\sqrt{xy}\)

=>2\(\sqrt{xy}-4\)≥\(0\)

<=> \(4-4\)≥0

<=>0≥0 ( Luôn đúng )

Vậy \(\dfrac{1}{x+3}+\dfrac{1}{y+3}\)≤\(\dfrac{2}{5}\)

\(\frac{-19}{6}+\frac{-15}{2}+\frac{11}{3}

tổng các số nguyên x là:

-5+-4+-3+-2+...+3+4

=-5+-4+4+-3+3+-2+2+...+0

=-5+0+0+0+...+0

=-5