3x2 + 6xy + 3y2 – 3z2

= 3.(x2 + 2xy + y2 – z2)

(Nhận thấy xuất hiện x2 + 2xy + y2 là hằng đẳng thức nên ta nhóm với nhau)

= 3[(x2 + 2xy + y2) – z2]

= 3[(x + y)2 – z2]

= 3(x + y – z)(x + y + z)

a) 10x(x-y) - 6y(y-x)
= 10x(x-y) +6y ( x-y)
=(10x+6y) (x-y)
b) 3x² + 5y - 3xy -5x
= 3x(x-y) + 5(y-x)
= 3x(x-y) -5(x-y)
= (3x-5) ( x-y)
c) 3y^₂ - 3z² +3x² + 6xy
=3(y² - z² + x² + 2xy)
=3[(x² +2xy+y²)-z² ]
=3[(x+y)² - z² ]
=3(x+y-z) (x+y+z)
d) 16x³ + 54y3
=2(8x³ + 27y³ )
=2[(2x)³ + (3y)³ ]
=2(2x+3y) (4x² - 6xy + 9y² )
e) x² - 25 -2xy+y²
=(x²-2xy+y²)-25
=(x-y)² -5²
=(x-y-5) (x-y+5)
f) (mình chưa làm ra )
{mong m.n bổ sung thêm..}

mấy câu trên bạn kia đã trả lời rồi nên mk k làm lại nx

f, x⁵ - 3x⁴ + 3x³ - x²

= x² (x³ - 3x² + 3x -1)

= x² (x - 1)³

Chúc bạn học tốt!

a) \(5x+10y=5\left(x+2y\right)\)

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

c) \(4x^2+9x-4xy-9y=4x\left(x-y\right)+9\left(x-y\right)=\left(x-y\right)\left(4x+9\right)\)

d) \(3x^2+5y-3xy-5x=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)

3x3 – 6x2 + 3x = 3x(x2 - 2x + 1) = 3x(x - 1)2

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

\(x^2-y^2-8y-16=x^2-\left(y^2+8y+16\right)=x^2-\left(y+4\right)^2=\left(x+y+4\right)\left(x-y-4\right)\)

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

\(3x^2-6xy+3y^2-27=3\left[\left(x-y\right)^2-9\right]=3\left(x-y-3\right)\left(x-y+3\right)\)

a. \(A=x^5-3x^3+x^2-x^3-3+2x=x^5-4x^3+x^2+2x-3\)

\(B=x^4-3x-2+5x^2-3x^4+2x^5=2x^5-2x^4+5x^2-3x-2\)

b. \(A+B=x^5-4x^3+x^2+2x-3+2x^5-2x^4+5x^2-3x-2\)

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

lũy thừa giảm dần hay tăng dần v ạ

a) \(2x^2+5x+2\)

\(=2x^2+4x+x+2\)

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

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

b) \(4x^2-4x-9y^2+12y-3\)

\(=\left(4x^2-4x+1\right)-\left(9y^2-12y+4\right)\)

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

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

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

c) \(x^4-2x^3-4x^2+4x-3\)

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

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

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

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

d) \(x^3-x+3x^2y+3xy^2+y^3-y\)

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

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

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

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

:) bóc lột !

Câu 1: 

a) 2x(3x+2) - 3x(2x+3) = 6x^2+4x - 6x^2-9x = -5x

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

\(=x^3+6x^2+12x+8+x^2-6x+9-x^3-5x^2\)

\(=2x^2+6x+17\)

c) \(\left(3x^3-4x^2+6x\right)\div\left(3x\right)=x^2-\dfrac{4}{3}x+2\)