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

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

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

x² - 6x - 4x + 24 = 0

( x² - 6x ) - ( 4x - 24 ) = 0

x( x - 6 ) - 4 ( x - 6 ) = 0

( x - 4 ) ( x - 6 ) = 0

\(\Rightarrow\orbr{\begin{cases}x-4=0\Rightarrow x=4\\x-6=0\Rightarrow x=6\end{cases}}\)

Vay x= 4 hoac x = 6

x² - 6x - 4x + 24 = 0

( x² - 6x ) - ( 4x - 24 ) = 0

x ( x - 6 ) - 4 ( x - 6 ) = 0

( x - 4 ) ( x - 6 ) = 0

\(\Rightarrow\orbr{\begin{cases}x-4=0\\x-6=0\end{cases}}\)

1. x - 4 = 0    => x = 4

2. x - 6 = 0    => x = 6

a: \(x^2+12x+36=0\) 

=>\(x^2+2\cdot x\cdot6+6^2=0\)

=>\(\left(x+6\right)^2=0\)

=>x+6=0

=>x=-6

b: \(4x^2-4x+1=0\)

=>\(\left(2x\right)^2-2\cdot2x\cdot1+1^2=0\)

=>\(\left(2x-1\right)^2=0\)

=>2x-1=0

=>2x=1

=>x=1/2

c: \(x^3+6x^2+12x+8=0\)

=>\(x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=0\)

=>\(\left(x+2\right)^3=0\)

=>x+2=0

=>x=-2

Bài 1 :

\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)

Bài 2 :

 \(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)

\(=x^6\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)+x^2+x+1-x^4\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)

=\(\left(x^2+x+1\right)\left(x^6+x^3+1-x^4-x\right)\)

Tick đúng nha 

X^2-6+8

\(3x^2-6x+3=0\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)

2.
a) 4x(x-1)-6x+6
= 4x(x-1)-6(x-1)
= (4x-6)(x-1)
3.
a) 6x²-24x=0
    6x(x-4)=0
TH1: 6x=0         TH2: x-4=0
           x=0                     x=4
Vậy x\(\in\){0;4}

2. a. \(4x\left(x-1\right)-6x+6\)

\(=4x\left(x-1\right)-6\left(x-1\right)\)

\(=\left(4x-6\right)\left(x-1\right)\)

3. a. \(6x^2-24x=0\)

\(\Leftrightarrow6x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}6x=0\\x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

đặt y=x²+1

=>y²=(x²+1)²

y²=x⁴+2x²+1

đặt P(x)=x^4+6x^3+11x^2+6x+1

=x⁴+2x²+1+6x³+6x+9x²

=x⁴+2x+1+6x(x²+1)+9x²

thay y²=x⁴+2x²+1 và y=x²+1 ta được 

Q(y)=y²+6xy+9x²

=(y+3x)²

thay y=x²+1 ta được:

(x²+3x+1)²

vậy x^4+6x^3+11x^2+6x+1=(x²+3x+1)²

Bài 1: \(0,125\left(a+1\right)^3-1\)

\(=\left(\frac12a+\frac12\right)^3-1\)

\(=\left(\frac12a+\frac12-1\right)\left\lbrack\left(\frac12a+\frac12\right)^2+\left(\frac12a+\frac12\right)+1^2\right\rbrack\)

\(=\left(\frac12a-\frac12\right)\left(\frac14a^2+\frac12a+\frac14+\frac12a+\frac12+1\right)=\left(\frac12a-\frac12\right)\left(\frac14a^2+a+\frac74\right)\)

BÀi 2:

\(28\cdot122-28\cdot62+30\cdot144\)

\(=28\cdot\left(122-62\right)+60\cdot72\)

\(=60\cdot\left(28+72\right)=60\cdot100=6000\)

Bài 3:

(6x+11)(5x-12)-42x+66=0

=>\(30x^2-72x+55x-132-42x+66=0\)

=>\(30x^2-59x-66=0\)

=>\(x^2-\frac{59}{30}x-\frac{66}{30}=0\)

=>\(x^2-2\cdot x\cdot\frac{59}{60}+\frac{3481}{3600}=\frac{11401}{3600}\)

=>\(\left(x-\frac{59}{60}\right)^2=\frac{11401}{3600}\)

=>\(\left[\begin{array}{l}x-\frac{59}{60}=\frac{\sqrt{11401}}{60}\\ x-\frac{59}{60}=-\frac{\sqrt{11401}}{60}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt{11401}+59}{60}\\ x=\frac{-\sqrt{11401}+59}{60}\end{array}\right.\)

\(=x\left(x-6\right)+2\left(x-6\right)=\left(x-6\right)\left(x+2\right)\)