1,\(=4x\left(x-\dfrac{3}{2}\right)\)
2,\(=-7y^3\left[2x^2y\left(2y+x\right)+3\right]\)
3, = 4x(a-b)-6xy(a-b)
=2x(a-b)(2-3y)
4,
=3(2x+1)-(2x-5)(2x+1)
=(3-2x+5)(2x+1)
=(8-2x)(2x+1)
=2(4-x)(2x+1)
 

5: \(4\left(x-3\right)^2+2x\left(3-x\right)\)

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

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

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

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

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

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

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

a)x^2-(a+b)x+ab

= x^2 - ax - bx + ab

= (x^2 - ax) - (bx - ab)

= x(x-a) - b(x-a)

= (x-b)(x-a) 

b)7x^3-3xyz-21x^2+9z

= 

c)4x+4y-x^2(x+y)

= 4(x + y) - x^2(x+y)

= (4-x^2) (x+y)

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

d) y^2+y-x^2+x

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

= (y-x)(y+x)+ (x+y)

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

e)4x^2-2x-y^2-y

= [(2x)^2 - y^2] - (2x +y)

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

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

f)9x^2-25y^2-6x+10y

= 

ko biết làm

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

a) \(3xy-6xy^2=3xy\left(1-2y\right)\)

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

c) \(x^3-x^2+2\)

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

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

f) \(x^2+2x+1-9y^2=\left(x+1\right)^2-\left(3y\right)^2=\left(x-3y+1\right)\left(x+3y+1\right)\)

g) \(6x^2-12x=6x\left(x-2\right)\)

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

i) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)

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

l) \(x^3-7x^2+9x+3x^2-21x+27=x\left(x^2-7x+9\right)+3\left(x^2-7x+9\right)=\left(x+3\right)\left(x^2-7x+9\right)\)

\(x^2+3x+2\)

\(=x^2+x+2x+2\)

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

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

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

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

\(=-5.\left(2x-1\right)\)

B) \(\left(4x^2+2xy+y^2\right)\left(2x-y\right)-\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)

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

\(=8x^3-y^3-8x^3-y^3\)

\(=-2y^3\)

C) \(x^2+6x+8\)

\(=x^2+6x+9-1\)

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

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

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

bài 3 A) \(x^2-16=0\)

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

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

vậy \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

B) \(x^4-2x^3+10x^2-20x=0\)

\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\left(x^3+10x\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^3+10x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x\left(x^2+10\right)=0\\x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

vậy \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

x=0

x=2

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

chuyển về dạng nguyên thể rồi tính thể chất khối lượng sau đó quay về đang tìm mũ của nhiều số làm ra rồi thì dễ lắm bạn ạ k minh nha

a)\(\left(x^2-2\right)\left(x^2+2x+2\right)\)

b)\(\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\)

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

d)\(\left(x-5\right)\left(4x+1\right)\)

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

g)\(2\left(a-b\right)^2\)

h)\(\left(xy-3\right)\left(5y^2-2z\right)\)

i)\(\left(4x+1\right)\left(2x-y\right)\)

l)\(abc^2\left(b-a\right)\left(b+c\right)\)

m)\(\left(x-y\right)\left(y-z\right)\left(x-z\right)\)