\(x^2+x=12\)

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

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

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

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

⇔\(x-\dfrac{1}{3}=-\dfrac{5}{6}\)

⇔\(x=-\dfrac{1}{2}\)

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

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

=>6x-24=-60

=>6x=-36

=x=-6

x : 1/3 = 1/2 +1/3

=>x:1/3=5/6

=>x=5/6 x 1/3

=>x=5/18

(x:2/3):2/5=7/10

=>x:2/3=7/10 x 2/5 

=>x:2/3=7/25

=>x=7/25x2/3

=>x=14/75

CHÚC BẠN HỌC TỐT

cái này vừa gửi rồi mà:D

x:1/3=1/2+1/3
x:1/3=5/6
x      =5/6.1/3
x      =5/18
 

(x:2/3):2/5=7/10
x:2/3        =7/10.2/5
x:2/3        =7/25
x:2/3        =7/25.2/3
x:2/3        =14/75
 

=>10x=9,78

hay x=0,978

\(x:33=\dfrac{9}{66}\)
\(x=\dfrac{9}{66}\times33\)
\(x=\dfrac{9}{2}\)
\(\dfrac{4}{5}\times x=\dfrac{10}{11}:\dfrac{3}{4}\)
\(\dfrac{4}{5}\times x=\dfrac{40}{33}\)
\(x=\dfrac{40}{33}:\dfrac{4}{5}\)
\(x=\dfrac{50}{33}\)
#Đạt Đang Bận Thở

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

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

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

a)x . 3,5 + x . 12,5 = 396,8
   x (3,5+12,5) =396,8 
   x. 16 =396,8
  x = 396,8 :16
 x= 24,8
b) 20,11 : 0,1 + 20,11: 10 -20,11 : (2,1 : 21) =2,011