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

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

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

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

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

\(25x^2-70xy+49y^2\\ =25x^2-35xy-35xy+49y^2\\ =25x\left(x-\dfrac{7}{5}y\right)-35y\left(x-\dfrac{7}{5}y\right)\\ =\left(25x-35y\right)\left(x-\dfrac{7}{5}y\right)\\ =5\left(5x-7y\right)\left(x-\dfrac{7}{5}y\right)\)

(5x-7y)^2

=(5x)^2-2*5x*7y+(7y)^2

=(5x-7y)^2

a) ( x 2  – 4x + 1)( x 2  – 2x + 3).

b) ( x 2  + 5x – 1)( x 2  + x – 1).

\(\left(25x^2-2\right)=\left(5x-\sqrt[]{2}\right)\left(5x+\sqrt[]{2}\right)\)

\(x^3-10x^2+25x-16xy^2\)

\(=x\left(x^2-10x+25-16y^2\right)\)

\(=x\left[\left(x-5\right)^2-16y^2\right]\)

\(=x\left(x-4y-5\right)\left(x+4y-5\right)\)

p/s: chúc bạn học tốt

a: \(=x^2\left(x-y\right)+2014\left(x-y\right)=\left(x-y\right)\left(x^2+2014\right)\)