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

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

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

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

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

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

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

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

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

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

b: \(\Leftrightarrow x+8\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{-7;-9;-3;-13\right\}\)

a)  218-(x+31)= -12

=> 218 + 12 - ( x + 31 ) = 0

=> 230 - x - 31 = 0

=> 199 - x = 0

=> x = 199

b)  \(\dfrac{x+5}{3}=\dfrac{6-x}{-2}\)

=> \(\dfrac{x+5}{3}=\dfrac{-6+x}{2}\)

=>2.( x + 5 ) = 3.( -6 + x )

=> 2x +10     = -18 +3x

=>        28         =  x

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.

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

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

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

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

giúp mình với nhé,mình đang cần gấp

\(a,\dfrac{6}{5}=\dfrac{18}{x}\\ \Rightarrow x=18:\dfrac{6}{5}\\ \Rightarrow x=15\\ b,\dfrac{3}{4}=\dfrac{-21}{x}\\ \Rightarrow x=-21:\dfrac{3}{4}\\ \Rightarrow x=-28\\ c,\dfrac{2}{-7}=\dfrac{18}{x}\\ \Rightarrow x=18:\dfrac{2}{-7}\\ \Rightarrow x=-63\\ d,\dfrac{-5}{2}=\dfrac{10}{-x}\\ \Rightarrow x=-10:\dfrac{-5}{2}\\ \Rightarrow x=4\)

\(a,\dfrac{6}{5}=\dfrac{18}{x}\Rightarrow6.x=5.18=90\\ \Rightarrow6.x=90\\ \Rightarrow x=15\\ b,\dfrac{3}{4}=\dfrac{-21}{x}\Rightarrow3.x=4.21=84\\ \Rightarrow x=28\)

\(a,\dfrac{6}{5}=\dfrac{18}{x}=>x=\dfrac{18.5}{6}=\dfrac{90}{6}=15\)

\(b,\dfrac{3}{4}=-\dfrac{21}{x}=>x=-\dfrac{21.4}{3}=-\dfrac{84}{3}=24\)

\(c,\dfrac{2}{-7}=\dfrac{18}{x}=>x=\dfrac{18.-7}{2}=-\dfrac{126}{2}=-63\)

\(d,-\dfrac{5}{2}=\dfrac{10}{-x}=>-x=\dfrac{10.2}{-5}=\dfrac{20}{-5}=-4=>x=4\)