\(a,\left(x+12\right)\left(x-6\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+12>0\\x-6>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+12< 0\\x-6< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-12\\x>6\end{matrix}\right.\\\left\{{}\begin{matrix}x< -12\\x< 6\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x>6\\x< -12\end{matrix}\right.\)

\(b,\left(10-x\right)\left(3-x\right)< 0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}10-x< 0\\3-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}10-x>0\\3-x< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>10\\x< 3\left(vô.lí\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x< 10\\x>3\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x< 10\\x>3\end{matrix}\right.\)

\(a,\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+12>0\\x-6>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+12< 0\\x-6< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>6\\x< -12\end{matrix}\right.\\ \Rightarrow x\in\left\{...;-15;-14;-13;7;8;9;...\right\}\\ b,\Rightarrow\left(x-10\right)\left(x-3\right)< 0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-10>0\\x-3< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-10< 0\\x-3>0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>10;x< 3\left(\text{loại}\right)\\3< x< 10\end{matrix}\right.\\ \Rightarrow x\in\left\{4;5;6;7;8;9\right\}\)

\(a,\left(x+12\right)\left(x+6\right)>0\) \(khi\) \(x>6\Rightarrow x\in\left\{7,8,9,...\right\}\)

\(b,\left(10-x\right)\left(3-x\right)< 0\) \(khi\) \(x< 10\Rightarrow x\in\left\{9,8,7,...\right\}\)

a -11

b 4

c 4

a) (Tớ đọc đề không hiểu, không nhớ :v)

b) a. `2(x-5) = 0`

`x-5  = 0:2`

`x-5 = 0`

`x=0+5`

`x=5`

b) `12(x-35) = 0`

`x-35=0:12`

`x-35 =0`

`x=0+35`

`x=35`

c) `(x-10).(x-13) = 0`

`=> x-10=0`

`x-13=0`

`=> x=0+10`

`x=0+13`

`=>x=10`

`x=13`.

a) Tích 2 số bằng không khi 1 trong 2 số bằng 0

b) \(2\left(x-5\right)=0\)

\(\Rightarrow x-5=0\)

\(\Rightarrow x=5\)

\(12\left(x-35\right)=0\)

\(\Rightarrow x-35=0\)

\(\Rightarrow x=35\)

\(\left(x-10\right)\left(x-13\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-10=0\\x-13=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=13\end{matrix}\right.\)

Lời giải:
a. Vì $x,y$ thuộc $Z$ nên $x-3, y+5\in\mathbb{Z}$. Tích của chúng $=11$ nên ta có bảng sau:

b. Vì $x,y\in\mathbb{Z}$ nên $2x+1, 6-y\in\mathbb{Z}$.

Với $x$ nguyên thì $2x+1$ là số nguyên lẻ nên ta có bảng sau:

a: =>-12x=60-12=48

=>x=-4

b: =>-5x=24+6-5=25

=>x=-5

Chi tiết nha

Lời giải:
a.

$(-12).x=(-15)(-4)-12$

$(-12)x=60-12=48$

$x=48:(-12)=-4$

b.

$(-5)x+5=(-3)(-8)+6$

$(-5)x+5=24+6=30$

$(-5)x=30-5=25$

$x=25:(-5)=-5$

a) \(5\times x-123=12\)

\(\Rightarrow5\times x=135\)

\(\Rightarrow x=27\)

b) \(x+3x+5x+7x=96\)

\(\Rightarrow16x=96\)

\(\Rightarrow x=6\)

a) \(5\times x-123=12\)

\(5x=12+123\)

\(5x=135\)

\(x=135:5\)

\(x=27\)

________

b) \(x+3x+5x+7x=96\)

\(x\left(1+3+5+7\right)=96\)

\(x.16=96\)

\(x=96:16\)

\(x=6\)

a)

\(5\times x-123=12\\\Rightarrow 5\times x=12+123\\\Rightarrow 5\times x=135\\\Rightarrow x=135:5\\\Rightarrow x=27\\---\)

b) 

\(x+3x+5x+7x=96\\\Rightarrow16x=96\\\Rightarrow x=96:16\\\Rightarrow x=6\)

\(a.\)

\(x-\dfrac{5}{6}=\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{1}{2}+\dfrac{5}{6}=\dfrac{3+5}{6}=\dfrac{8}{6}=\dfrac{4}{3}\)

\(b.\)

\(-\dfrac{3}{4}-x=-\dfrac{7}{12}\)

\(\Leftrightarrow x=-\dfrac{3}{4}--\dfrac{7}{12}=-\dfrac{3}{4}+\dfrac{7}{12}=\dfrac{\left(-3\right)\cdot3+7}{12}=\dfrac{-2}{12}=-\dfrac{1}{6}\)

a)x=1/2+5/6

x=/3/6+5/6

x=8/6=4/3

b)x=-3/4- (-7/12)

x=-9/12-(-7/12)

x=-1/6

a) \(x-\dfrac{5}{6}=\dfrac{1}{2}\) 

           \(x=\dfrac{1}{2}+\dfrac{5}{6}\) 

          \(x=\dfrac{4}{3}\) 

b) \(\dfrac{-3}{4}-x=\dfrac{-7}{12}\) 

               \(x=\dfrac{-3}{4}-\dfrac{-7}{12}\) 

               \(x=\dfrac{-1}{6}\)

a: x:(-9)=-54

=>\(x=\left(-54\right)\cdot\left(-9\right)\)

=>\(x=54\cdot9=486\)

b: \(x:\left(-12\right)=18\)

=>\(x=18\cdot\left(-12\right)=-216\)

c: \(x:\left(-5\right)=-19\)

=>\(x:5=19\)

=>\(x=19\cdot5=95\)

d: \(\left(x-28\right):\left(-12\right)=-5\)

=>\(x-28=\left(-12\right)\cdot\left(-5\right)=60\)

=>x=60+28=88

e: \(\left(x+15\right):\left(-28\right)=8\)

=>x+15=-28*8=-224

=>x=-224-15=-239

f: (x+30):(-45)=-4

=>\(x+30=\left(-45\right)\cdot\left(-4\right)=180\)

=>x=180-30

=>x=150

a) x : (-9) = -54

x= -54 . (-9)= 486

________

b) x : (-12) = 18

x= 18. (-12)= -216

_________

c) x : (-5) = -19

x= (-19). (-5)= 95

__________

d) (x - 28) : (-12) = -5

(x-28)= (-5). (-12)= 60

x= 60+28= 88

_______

e) (x + 15) : (-28) = 8

(x+15)= 8. (-28)= -224

x= -224 - 15 = - 239

__________

f) (x + 30) : (-45) = -4

(x+30)= -4. (-45)= 180

x= 180 - 30=150

a) Ta có: 70 ⋮ x và 84 ⋮ x 

Nên \(x\inƯC\left(70;84\right)\)

\(Ư\left(70\right)=\left\{\text{1,2,7,10,14,35,70}\right\}\) 

\(Ư\left(84\right)=\left\{\text{ 1 ; 2 ; 3 ; 4 ; 6 ; 7 ; 12 ; 14 ; 21 ; 28 ; 42 ; 84 }\right\}\)

Mà: \(x>8\)

\(\Rightarrow x\in\left\{14\right\}\)

b) Ta có: x ⋮ 12 và x ⋮ 15 nên \(\Rightarrow x\in BC\left(12;15\right)\)

\(B\left(12\right)=\left\{0;...;84;96;108;120;132;144;156;...\right\}\)

\(B\left(15\right)=\left\{0;...;90;105;120;135;150;165;180;195;...\right\}\)

Mà: \(100< x< 300\)

\(\Rightarrow x\in\left\{120;180;240\right\}\)

a) Do 70 ⋮ x và 84 ⋮ x nên x là ước chung của 70 và 84

Ta có:

70 = 2.5.7

84 = 2².3.7

⇒ ƯCLN(70; 84) = 2.7 = 14

⇒ ƯC(70; 84) = {1; 2; 7; 14}

Do x > 8 nên x = 14

b) Do x ⋮ 12 và x ⋮ 15 nên x là bội chung của 12 và 15

Ta có:

12 = 2².3

15 = 3.5

⇒ BCNN(12; 15) = 2².3.5 = 60

⇒ BC(12; 15) = B(60) = {0; 60; 120; 180; 240; 300;...}

Do 100 < x < 300 nên x ∈ {120; 180; 240}

a,-4/7=x/21
-12/21 = x/21
x= -12 

b,(x-3)/15=1/-5
x - 3 = -1/5 * 15
x - 3 = -3
x = 0

c,.(3x+8)/-12=-5/30
=> 3x + 8 = 2
=> 3x=-6
=>x=-2