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

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

⇔\(x-\dfrac{1}{3}=-\dfrac{5}{6}\)

⇔\(x=-\dfrac{1}{2}\)

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

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

=>6x-24=-60

=>6x=-36

=x=-6

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: 

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

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

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

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

x/30 + 4/5= 79/30

x/30 = 79/30 - 4/5

x/30= 11/6

11/6= 55/30

=> x= 55

x^3+1=9

x^3=8

x^3=2^3

=>x=2

=>x³=8

hay x=2

`x^3+1=9`

`x^3=8`

`x^3=2^3`

`=>x=2`

bài này mình biết làm nhưng thôi

\(\frac{x+1}{2}=\frac{8}{x+1}\Rightarrow\left(x+1\right)^2=8.2\)

=>(x+1)²=16

=>x+1=\(\sqrt{16}\)

=>x+1=4

=>x=3

`x-1/8=1/4xx2/3`

`x-1/8=1/6`

`x=1/6+1/8`

`x=7/24`

x - 1/8 = 1/6

x         = 6/48 + 8/48
x         = 14/48

x - 1/8 = 1/4 x 2/3

x - 1/8=   1/6

x        = 1/6 + 1/8

x       =  14/48