\(a,\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-5< 0\\x+2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-5>0\\x+2< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 5\\x>-2\end{matrix}\right.\\\left\{{}\begin{matrix}x>5\\x< -2\end{matrix}\right.\end{matrix}\right.\Rightarrow-2< x< 5\\ \Rightarrow x\in\left\{-1;0;1;2;3;4\right\}\\ b,\Rightarrow5< x^2< 14\\ \Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

(2-x)^3+(2+x)^3-12x(x+1)=0

=>\(8-12x+6x^2-x^3+8+12x+6x^2+x^3-12x\left(x+1\right)=0\)

=>\(12x^2+16-12x^2-12x=0\)

=>16-12x=0

=>4-3x=0

=>x=4/3

Lời giải:

a. $x^2-100x=0$

$\Leftrightarrow x(x-100)=0$

$\Rightarrow x=0$ hoặc $x-100=0$

$\Leftrightarrow x=0$ hoặc $x=100$

b.

$x^2+5x+6=0$

$\Leftrightarrow (x^2+2x)+(3x+6)=0$

$\Leftrightarrow x(x+2)+3(x+2)=0$

$\Leftrightarrow (x+2)(x+3)=0$

$\Leftrightarrow x+2=0$ hoặc $x+3=0$

$\Leftrightarrow x=-2$ hoặc $x=-3$

a) x = 0 :))

TH1:x=1

TH2:x=2

TH3:x=3

\(0\le x-1\le2\)

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

\(\Rightarrow x\in\left\{1;2;3\right\}\)

Vậy \(x\in\left\{1;2;3\right\}\)

a) 0 < x-1 < 2

=> x-1 E Ư{0;1;2}

=> x E Ư{1;2;3}

Vậy x E Ư[1;2;3}

a ,\(4x^2-\left(x-3\right)^2=0\)

\(\Leftrightarrow\left(2x-x+3\right)\left(2x+x-3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(3x-3\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+3=0\\3x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\3x=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

Vậy 

b,\(x^2-4+\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x+2\right)^2=0\)

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

Vậy ...

\(a,x\left(x+5\right)-\left(x-2\right)\left(x+3\right)=0\\ \Leftrightarrow x^2+5x-x^2-x+6=0\Leftrightarrow4x=-6\\ \Leftrightarrow x=-\dfrac{3}{2}\)

\(b,2x^3-18x=0\\ \Leftrightarrow2x\left(x^2-9\right)=0\\ \Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

a: Ta có: \(x\left(x+5\right)-\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow x^2+5x-x^2-3x+2x+6=0\)

\(\Leftrightarrow7x=-6\)

hay \(x=-\dfrac{6}{7}\)

b: Ta có: \(2x^3-18x=0\)

\(\Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

\(a,\Leftrightarrow\left(x+3\right)\left(x+3-x+3\right)=0\Leftrightarrow x=-3\\ b,\Leftrightarrow x=0\left(x^2+4>0\right)\)

\(a,x^2+2.x.3+3^2-\left(x^2-3^2\right)=0\)

\(x^2+6x+9-x^2+9=0\)

\(6x+18=0\)

\(6x=-18\)

\(x=-3\)

Vậy x=-3

\(b,5x^3+20x=0\)

\(5x\left(x^2+4\right)=0\)

\(Th1:5x=0=>x=0\)

\(Th2:x^2+4=0\)

\(x^2=-4\)(vô lý)

Vậy x=0