Bài 1: 

a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)

\(=\left(2x+1\right)\left(3-2x+5\right)\)

\(=\left(2x+1\right)\left(8-2x\right)\)

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

b) Ta có: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)

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

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

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

\(=3\left(3x-2\right)\left(x-2\right)\)

Bài 2: 

a: Ta có: \(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)

\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)

\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)

\(=\left(a-b\right)\left(2a-4b\right)\)

\(=2\left(a-b\right)\left(a-2b\right)\)

f: Ta có: \(x^2-6xy+9y^2+4x-12y\)

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

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

whatterfice

B1

A=11x^2-x-2

B=2(-4+x)

B2

a)=(x+3)^2(x-3)

B2

b)=(5x-7)(x+2)

\(a,=2x\left(x+3\right)\\ b,=x^3\left(x+3\right)+\left(x+3\right)=\left(x^3+1\right)\left(x+3\right)\\ =\left(x+1\right)\left(x+3\right)\left(x^2-x+1\right)\\ c,=64-\left(x-y\right)^2=\left(8-x+y\right)\left(8+x-y\right)\\ A=x^2+6x+5+x^3-8-x^2-x+2\\ A=x^3+5x-1\)

a) 2x²+6x=2x(x+3)
b) x⁴+3x³+x+3=(x⁴+x)+(3x³+3)=x(x³+1)+3(x³+1)=(x+3)(x³+1)
c) 64-x²-y²+2xy=-(x²-2xy+y²)+8²=8-(x+y)²=(8+x+y)(8-x-y)

A= (x+5)(x+1)+(x-2)(x²+2xx+4)-(x²+x-2)
A= x²+6x+5+x³-8-x²-x+2
A= x³+(x²-x²)+(6x-x)+(5-8+2)
A= x³+5x-1

Cưu là mình vs (x^2+x)^2-2(x^2+x)-15

2:

a: \(x^2-12x+20\)

\(=x^2-2x-10x+20\)

=x(x-2)-10(x-2)

=(x-2)(x-10)

b: \(2x^2-x-15\)

=2x^2-6x+5x-15

=2x(x-3)+5(x-3)

=(x-3)(2x+5)

c: \(x^3-x^2+x-1\)

=x^2(x-1)+(x-1)

=(x-1)(x^2+1)

d: \(2x^3-5x-6\)

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

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

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

e: \(4y^4+1\)

\(=4y^4+4y^2+1-4y^2\)

\(=\left(2y^2+1\right)^2-\left(2y\right)^2\)

\(=\left(2y^2+1-2y\right)\left(2y^2+1+2y\right)\)

f; \(x^7+x^5+x^3\)

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

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

\(=x^3\left[\left(x^2+1\right)^2-x^2\right]\)

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

g: \(\left(x^2+x\right)^2-5\left(x^2+x\right)+6\)

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

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

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

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

h: \(\left(x^2+2x\right)^2-2\left(x+1\right)^2-1\)

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

\(=\left[\left(x+1\right)^2-1\right]^2-2\left(x+1\right)^2-1\)

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

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

\(=\left(x+1\right)^2\left[\left(x+1\right)^2-4\right]\)

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

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

i: \(x^2+4xy+4y^2-4\left(x+2y\right)+3\)

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

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

\(=\left(x+2y\right)\left(x+2y-1\right)-3\left(x+2y-1\right)\)

\(=\left(x+2y-1\right)\left(x+2y-3\right)\)

j: \(x\cdot\left(x+1\right)\left(x+2\right)\left(x+3\right)-3\)

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

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

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

Bài 2

a) 5x² + 30y

= 5(x² + 6y)

b) x³ - 2x² - 4xy² + x

= x(x² - 2x - 4y² + 1)

= x[(x² - 2x + 1) - 4y²]

= x[(x - 1)² - (2y)²]

= x(x - 1 - 2y)(x - 1 + 2y)

Bài 3:

a: \(2x\left(x-3\right)-x+3=0\)

=>\(2x\left(x-3\right)-\left(x-3\right)=0\)

=>(x-3)(2x-1)=0

=>\(\left[{}\begin{matrix}x-3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)

b: \(\left(3x-1\right)\left(2x+1\right)-\left(x+1\right)^2=5x^2\)

=>\(6x^2+3x-2x-1-x^2-2x-1=5x^2\)

=>\(5x^2-x-2=5x^2\)

=>-x-2=0

=>-x=2

=>x=-2

\(B1\\ a,2x+10y=2\left(x+5y\right)\\ b,x^2+4x+4=x^2+2.2x+2^2=\left(x+2\right)^2\\ c,x^2-y^2+10y-25\\ =\left(x^2-y^2\right)+5\left(2y-5\right)\\ =\left(x-y\right)\left(x+y\right)+5\left(2y-5\right)\\ B2\)

\(a,x^2-3x+x-3=0\\ =>x\left(x-3\right)+\left(x-3\right)=0\\ =>\left(x+1\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ b,2x\left(x-3\right)-\dfrac{1}{2}\left(4x^2-3\right)=0\\ =>2x^2-6x-2x^2+\dfrac{3}{2}=0\\ =>-6x=-\dfrac{3}{2}\\ =>x=\left(-\dfrac{3}{2}\right):\left(-6\right)\\ =>x=\dfrac{1}{4}\\ c,x^2-\left(x-3\right)\left(2x-5\right)=9\\ =>x^2-2x^2+6x+5x-15=9\\ =>-x^2+11-15-9=0\\ =>-x^2+11x-24=0\\ =>-x^2+8x+3x-24=0\\ =>-x\left(x-8\right)+3\left(x-8\right)=0\\ =>\left(3-x\right)\left(x-8\right)=0\\ =>\left[{}\begin{matrix}3-x=0\\x-8=0\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=8\end{matrix}\right.\)

Bài 2: 

a: \(3x^2-3xy=3x\left(x-y\right)\)

b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)

c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)

d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)

ỳtct7ct7c7c7t79tc9

a) \(=x^4-14x^2+40-72=x^4-14x^2-32=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)

b) \(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1=\left(x^2+5x\right)^2+2\left(x^2+5x\right)+1=\left(x^2+5x+1\right)^2\)

c) \(=x^4+3x^3-3x^2+3x^3+9x^2-9x+x^2+3x-3-5=x^4+6x^3+7x^2-6x-8=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)

a: Ta có: \(\left(x^2-4\right)\left(x^2-10\right)-72\)

\(=x^4-14x^2-32\)

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

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

b: Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)

\(=\left(x^2+5x+6\right)\left(x^2+5x+4\right)+1\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24+1\)

\(=\left(x^2+5x+1\right)^2\)