\(\left|x-3\right|\inƯ\left(13\right)\Leftrightarrow\left|x-3\right|\in\left\{\pm1;\pm13\right\}\)

Th1: \(\left|x-3\right|=1\)\(\Leftrightarrow\left[\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=4\\x=2\end{matrix}\right.\)

Th2: \(\left|x-3\right|=-1\Leftrightarrow x\in\varnothing\)vì \(\left|x-3\right|\ge0\)

Th3: \(\left|x-3\right|=13\Leftrightarrow\left[\begin{matrix}x-3=13\\x-3=-13\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=16\\x=-10\end{matrix}\right.\)

Th4: \(\left|x-3\right|=-13\Leftrightarrow x\in\varnothing\)vì \(\left|x-3\right|\ge0\)

Vậy \(x\left\{2;4;-10;16\right\}\)

\(17+\left|-23\right|-\left(-40\right)=17+23+40=80\)

\(\frac{2n+15}{n+1}\in Z\Leftrightarrow2n+15⋮n+1\Leftrightarrow2n+2+13⋮n+1\Leftrightarrow2\left(n+1\right)+13⋮n+1\)\(\Leftrightarrow13⋮n+1\) \(\left(vì2\left(n+1\right)⋮n+1\right)\)

\(\Leftrightarrow n+1\inƯ\left(13\right)\Leftrightarrow n+1\in\left\{\pm1;\pm13\right\}\Leftrightarrow n\in\left\{0;-2;12;-14\right\}\)

Vậy\(n\in\left\{0;-2;12;-14\right\}\)

a) Vì OA < OB nên A nằm giữa O và B

b) Vì O nằm giữa A và B nên OA + AB = OB

\(\Rightarrow\) AB = OB - OA = 4 - 2 = 2 (cm) (1)

Vì OA < OC nên A nằm giữa O và C

\(\Rightarrow\) OA + AC = OC

\(\Rightarrow\) AC = OC - OA = 7 - 2 = 5 (cm) (2)

Từ (1);(2) \(\Rightarrow\) AB < AC (vì 2 < 5) \(\Rightarrow\) B nằm giữa A và C

Ta có giá trị tuyệt đối của một số luôn lớn hơn hoặc bằng 0

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

Mà x=2=>x=-y<=>2=-y<=>y=-2

=>-2=2z<=>z=-1

Vậy: x=2; y=-2; z=-1

-279.(-63-179)-179.(279+63)

=279.63+279.179-179.279-179.63

=(279.63-179.63)+(279.179-179.279)

=63.(279-179)+0

=63.100=6300

\(-5⋮n-2\Leftrightarrow n-2\inƯ\left(5\right)\Leftrightarrow n-2\in\left\{\pm1;\pm5\right\}\Leftrightarrow n-2\in\left\{1;3;-3;7\right\}\)

Vậy \(n\in\left\{1;3;-3;7\right\}\)

5x+9=2x+15<=>5x-2x=15-9<=>3x=6<=>x=3

(x-5).(x+4)=0

Th1x-5=0<=>x=5

Th2x+4=0<=>x=-4

Vậy x=0;-4

2.(x-3)-4(x+4)=3.(-7)+5<=>2x-6-4x-16=-21+5<=>-2x-22=-16<=>-2x=-16+22

<=>-2x=6<=>2x=-6<=>x=-3

/-15/-/x/=/-12/

Th1:x<0=>/-15/-/x/=/-12/<=>15-(-x)=12<=>15+x=12<=>x=27

Th2:x> hoặc= 0=>/-15/-/x/=/-12/<=>15-x=12<=>15-12=x<=>x=3

3x-(-3)=7<=> 3x+3=7<=>3x=7-3<=>3x=4<=>x=4/3

\(2\left|x-20\right|=10\Leftrightarrow\left|x-20\right|=5\Leftrightarrow\left[\begin{matrix}x-20=5\\x-20=-5\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=25\\x=15\end{matrix}\right.\)

Vậy\(x\in\left\{25;15\right\}\)

\(2.\left|x-5\right|=2^5\Leftrightarrow2.\left|x-5\right|=32\Leftrightarrow\left|x-5\right|=16\Leftrightarrow\left[\begin{matrix}x-5=16\\x-5=-16\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x=21\\x=-11\end{matrix}\right.\)Vậy\(x\in\left\{-11;21\right\}\)