b: -7<x<7

a, \(\Rightarrow x-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)

b, \(3\left(x-2\right)+13⋮x-2\Rightarrow x-2\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)

c, \(x\left(x+7\right)+2⋮x+7\Rightarrow x+7\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

\(a,\Leftrightarrow x\left(x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\\ b,\Leftrightarrow\left(x+4-4\right)\left(x+4+4\right)=0\\ \Leftrightarrow x\left(x+8\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\\ c,\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\)

a) \(\Leftrightarrow x\left(x+9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\)

b) \(\Leftrightarrow x\left(x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)

c) \(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

d) \(\Leftrightarrow\left(x-5\right)^2=0\\ \Leftrightarrow x=5\)

a: \(\Leftrightarrow x-3=-13\)

hay x=-10

\(x^2=0\)

\(\Rightarrow x^2=0^2\)

\(\Rightarrow x=0\)

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\(x^2=16\)

\(\Rightarrow x^2=\left(\pm4\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x^2=\left(-4\right)^2\\x^2=4^2\end{matrix}\right.\)

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

\(x^2=0\)

\(\Rightarrow x=0\left(tm\right)\)

Vậy: \(x=0.\)

\(---\)

\(x^2=16\)

\(\Rightarrow x^2=\left(\pm4\right)^2\)

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

Vậy: \(x\in\left\{4;-4\right\}.\)

a) Khi m = 0 thì phương trình trở thành:

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

\(\Leftrightarrow x^2+2\cdot-2x-0=0\)

\(\Leftrightarrow x^2-4x=0\)

\(\Leftrightarrow x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

b) Ta có: 

\(\left|x_1\right|-\left|x_2\right|=6\)

\(\Leftrightarrow x^2_1+x_2^2-2\left|x_1x_2\right|=36\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2-2\left|x_1x_2\right|=36\)

Mà: \(x_1+x_2=-2\left(m-2\right)=4-2m\)

\(x_1x_2=-m^2\)

\(\Leftrightarrow\left(4-2m\right)^2-2\cdot-m^2-2\cdot m^2=36\)

\(\Leftrightarrow16-16m+4m^2+2m^2-2m^2=36\)

\(\Leftrightarrow\left(4-2m\right)^2=6^2\)

\(\Leftrightarrow\left[{}\begin{matrix}4-2m=6\\4-2m=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2m=-2\\2m=10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}m=-1\\m=5\end{matrix}\right.\)

a: \(x^2\left(x^2-9\right)=0\)

=>\(\left[\begin{array}{l}x^2=0\\ x^2-9=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^2=9\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=3\\ x=-3\end{array}\right.\)

b: \(2x^2-x-6=0\)

=>\(2x^2-4x+3x-6=0\)

=>2x(x-2)+3(x-2)=0

=>(x-2)(2x+3)=0

=>\(\left[\begin{array}{l}x-2=0\\ 2x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=-\frac32\end{array}\right.\)

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

\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

b: Ta có: \(x^2\left(x-6\right)-x^2+36=0\)

\(\Leftrightarrow\left(x-6\right)\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x-3\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=3\\x=-2\end{matrix}\right.\)