Ta có:

\(\left(x+5\right)\left(3x-12\right)=\left(x+5\right)3\left(x-4\right)=3.\left[\left(x+5\right)\left(x-4\right)\right]\)

Để \(3.\left[\left(x+5\right)\left(x-4\right)\right]<0\) thì x+5 và x-4 trái dấu.

Mà x+5>x-4 

\(\Rightarrow x+5>0\) và \(x-4<0\)

\(\Rightarrow x>-5\) và \(x<4\)

x là số nguyên ta có \(x\in\left\{-4;-3;-2;-1;1;2;3\right\}\)

Vậy \(x\in\left\{-4;-3;-2;-1;1;2;3\right\}\)

X=1 , X=2 , X=3

a) \(\left(x-1\right)\left(x^3+8\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\x^3+8=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x^3=-8\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

b) \(\left(x+1\right)\left(2x^2-8\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\2x^2-8=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x^2=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=\pm2\end{matrix}\right.\)

c) Vì : \(x^2+3\ge3>0\forall x\)

nên để : \(\left(x^2+3\right)\left(x+5\right)< 0\)

Thì : \(x+5< 0\\ \Rightarrow x< -5\)

x-1)(x-2)=0

⇒\(\left\{{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Ta có các trường hợp:

+TH1: \(\left\{{}\begin{matrix}x+2>0\\x-1>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x>-2\\x>1\end{matrix}\right.\)\(\Leftrightarrow x>1\)

+TH2: \(\left\{{}\begin{matrix}x+2< 0\\x-1< 0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x< -2\\x< 1\end{matrix}\right.\)\(\Leftrightarrow x< -2\)

Vậy.....

(x+2) (x-1)>0 thì nó có cả đống bạn ạ VD:

(10+2)x(11-1)= 120 > 0

Ta có các trường hợp sau:

+TH1: \(\left\{{}\begin{matrix}x+3>0\\x-2< 0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x>-3\\x< 2\end{matrix}\right.\)\(\Leftrightarrow-3< x< 2\)

+TH2: \(\left\{{}\begin{matrix}x+3< 0\\x-2>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x< -3\\x>2\end{matrix}\right.\) (vô lý)

Vậy -3<x<2

a, 12 - (2\(x^2\) - 3) = 7

            2\(x^2\)  - 3  =  12  - 7

           2\(x^2\) - 3  = 5

           2\(x^2\)  = 8

             \(x^2\)   = 4

             \(\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)

a) \(12-\left(2x^2-3\right)=7\\ 12-2x^2+3=7\\ 15-2x^2=7\\ 2x^2=15-7=8\\ x^2=8:2=4\\ x=\pm2\)

b) \(3x^2-12=2x^2+4\\ 3x^2-2x^2=12+4\\ x^2=16\\ x=\pm4\)

b, 3\(x^2\) - 12  = 2\(x^2\) + 4

    3\(x^2\) - 2\(x^2\) = 12 + 4

     \(x^2\) = 16

      \(\left[{}\begin{matrix}x=-4\\x=4\end{matrix}\right.\)

ta có x + y = 2 suy ra x= 2 - y

z + x = -5 suy ra x= -5-z

suy ra x=2 -y = -5 -z=-5-z-2+y= -7 - z + y

thay x=-7 - z + y vào z + x = -5 ta được 

z - 7 -z +y = - 5

-7 + y = -5

y=2

suy ra x= -2 , z=-3