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
 

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

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

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

\(x^2+x=12\)

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

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

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

x = 22 x 7/11

x = 14

3/2 x X = 11/2 - 7/4

3/2 x X = 22/7

X = 22/7 : 3/2

X = 44/21

\(\Leftrightarrow\left(x+2\right)^2\cdot\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-1\end{matrix}\right.\)

\(\Leftrightarrow x^3+5x^2+8x+4=0\\ \Leftrightarrow\left(x+1\right)\left(x+2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)

X=4/15:2/5=2/3

2/3

X = 4/5 : 2/15

X = 2/3

\(\Leftrightarrow x^3+2x^2-3x-x^3-3x^2=-4\)

\(\Leftrightarrow x^2+3x-4=0\)

=>(x+4)(x-1)=0

=>x=-4 hoặc x=1