\(x-\dfrac{4}{12}=-\dfrac{5}{6}\)

⇔\(x-\dfrac{1}{3}=-\dfrac{5}{6}\)

⇔\(x=-\dfrac{1}{2}\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

=>(x-4).6=12.(-5)

=>6x-24=-60

=>6x=-36

=x=-6

\(x^2+x=12\)

\(\Rightarrow x^2+x-12=0\)

\(\Rightarrow x\left(x-3\right)+4\left(x-3\right)=0\Rightarrow\left(x-3\right)\left(x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)

Ta có x*6*8=96

=>x*48=96

=>x    =96:48

=>x    =2

\(\dfrac{x}{5}=\dfrac{5}{6}-\dfrac{19}{30}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{25}{30}-\dfrac{19}{30}=\dfrac{6}{30}=\dfrac{1}{5}\)

\(\Leftrightarrow x=1\)

Ta có : \(\dfrac{x}{5}=\dfrac{5}{6}+\left(-\dfrac{19}{30}\right)=\dfrac{25}{30}-\dfrac{19}{30}=\dfrac{1}{5}\)

\(\Leftrightarrow5x=5\)

\(\Leftrightarrow x=1\)

Vậy ...

Ta có: \(\dfrac{x}{5}=\dfrac{5}{6}-\dfrac{19}{30}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{25}{30}-\dfrac{19}{30}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{6}{30}=\dfrac{1}{5}\)

hay x=1

Vậy: x=1

Bài 1: 

\(=\dfrac{-3-39}{42}+\dfrac{-6-11}{17}-\dfrac{1}{6}=-\dfrac{119}{48}\)

Bài 2: 

=>x:5=-13/20

hay x=-65/20=-13/4

x/30 + 4/5= 79/30

x/30 = 79/30 - 4/5

x/30= 11/6

11/6= 55/30

=> x= 55

\(x-\frac{x}{3}=\frac{3}{57}:\frac{12}{19}\)

\(x-\frac{x}{3}=\frac{3}{57}\times\frac{19}{12}\)

\(x-\frac{x}{3}=\frac{1}{12}\)

\(\Rightarrow12x-4x=1\)

\(\Rightarrow8x=1\)

\(\Rightarrow x=\frac{1}{8}\)

Bài 1: 

\(=\dfrac{-3-39}{32}+\dfrac{-6-11}{17}+\dfrac{-1}{6}=-\dfrac{21}{16}+\dfrac{-1}{6}-1=-\dfrac{119}{48}\)

Bài 2: 

\(\Leftrightarrow x:5=-\dfrac{13}{20}\)

hay x=-13/4

⇔x:5=−1320⇔x:5=−1320

hay x=-13/4

1. 

\(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\\ =\left(12x^2+6x\right)\left(y+z+y-z\right)\\ =2y\left(12x^2+6x\right)\\ =2y.6x\left(2x+1\right)\\ =12xy\left(2x+1\right)\)

2. 

\(x\left(x-6\right)+10\left(x-6\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)

Vậy \(x\in\left\{6;-10\right\}\) là nghiệm của pt

Bài 1:

Ta có: \(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\)

\(=\left(12x^2+6x\right)\left(y+z+y-z\right)\)

\(=6x\left(2x+1\right)\cdot2y\)

\(=12xy\left(2x+1\right)\)

Bài 2: 

Ta có: \(x\left(x-6\right)+10\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)