\(=-12x^2\left(y-x\right)+18x^3\left(y-x\right)\)

\(=-6x^2\left(y-x\right)\left(2-3x\right)\)

x²-7x+12

=x²-4x-3x+12

=x(x-4)-3(x-4)

=(x-3)(x-4)

        \(x^2-7x+12\)

\(=\left(x^2-4x\right)-\left(3x-12\right)\)

\(=x\left(x-4\right)-3\left(x-4\right)\)

\(=\left(x-4\right)\left(x-3\right)\)

c ) \(x^2-7x+12\)

\(=\left(x^2-3x\right)-\left(4x-12\right)\)

\(=x\left(x-3\right)-4\left(x-3\right)\)

\(=\left(x-4\right)\left(x-3\right)\)

d ) \(x^2+7x+12\)

\(=\left(x^2+3x\right)+\left(4x+12\right)\)

\(=x\left(x+3\right)+4\left(x+3\right)\)

\(=\left(x+4\right)\left(x+3\right)\)

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 2  – 4x + 1)( x 2  – 2x + 3).

b) ( x 2  + 5x – 1)( x 2  + x – 1).

12x²+7x-12=-7

Đúng 100% 

Mình mới lớp 6 nhưng mình nghĩ thế này nha :

x(12x +7) -12

phân tích cơ mak

(x^2+3x+2)(x^2+7x+12)

=(x²+x+2x+2)(x²+3x+4x+12)

=[x.(x+1)+2.(x+1)][x.(x+3)+4.(x+3)]

=(x+1)(x+2)(x+3)(x+4)