l/ $6x^2(x-1)-9x(x-1)\\=(6x^2-9)(x-1)\\=3(2x^2-3)(x-1)\\=3(\sqrt2 x-\sqrt 3)(\sqrt 2 x+\sqrt 3)(x-1)$

m/ $4x^2(x-2)+9x(2-x)\\=4x^2(x-2)-9x(x-2)\\=(4x^2-9x)(x-2)\\=x(4x-9)(x-2)$

n/ $4x^2y-4xy+y\\=y(4x^2-4x+1)\\=y(2x-1)^2$

o/ $3x(2x-3y)-6(3y-2x)\\=3x(2x-3y)+6(2x-3y)\\=(3x+6)(2x-3y)\\=3(x+2)(2x-3y)$

p/ $4x^2(x-1)+(1-x)\\=4x^2(x-1)-(x-1)\\=(4x^2-1)(x-1)\\=(2x-1)(2x+1)(x-1)$

l)\(6x^2\left(x-1\right)-9x\left(x-1\right)=3x\left(x-1\right)\left(2x-3\right)\)

m) \(4x^2\left(x-2\right)+9x\left(2-x\right)=4x^2\left(x-2\right)-9x\left(x-2\right)=x\left(x-2\right)\left(4x-9\right)\)

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

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

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

a) A=5xⁿy³ chia hết cho B=4x³y

ta có:

5xⁿy³ : 4x³y = \(\dfrac{5}{4}\) x ^n-3 y²

để A \(⋮\) B thì : n - 3 \(\ge\) 0

n \(\ge\) 3

1A:

a: \(x^3+2x=x\left(x^2+2\right)\)

b: \(3x-6y=3\left(x-2y\right)\)

c: \(5\left(x+3y\right)-15x\left(x+3y\right)\)

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

d: \(3\left(x-y\right)-5x\left(y-x\right)\)

\(=3\left(x-y\right)+5x\left(x-y\right)\)

\(=\left(x-y\right)\left(5x+3\right)\)

1A. a. x(x²+2) 

b. 3(x-2y)

c. 5(x+3y)(1-3x) 

d. (x-y) (3-5x)

1B. a. 2x(2x-3)

b.xy(x²-2xy+5)

c. 2x(x+1)(x+2)

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

1B:

a: \(4x^2-6x=2x\left(2x-3\right)\)

b: \(x^3y-2x^2y^2+5xy\)

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

Ta có f(x) + g(x) = 4x-2.

Cho 4x - 2 = 0 ⇒ 4x = 2 ⇒ x = 1/2. Chọn A

a) Ta có: \(P=-x^3y-xy+x^2+4x^3y+2xy+1\)

\(=3x^3y+xy+x^2+1\)

Bậc của đa thức P là 4

Ta có: \(Q=x^3y-8xy-5+2x^3y+9x^2+4-10x^2\)

\(=3x^3y-8xy-x^2-1\)

Bậc của đa thức Q là 4

b) Ta có: A=P+Q

\(=3x^3y+xy+x^2+1+3x^3y-8xy-x^2-1\)

\(=6x^3y-7xy\)

Ta có: B=P-Q

\(=3x^3y+xy+x^2+1-3x^3y+8xy+x^2+1\)

\(=9xy+2x^2+2\)

c) Thay x=1 và y=-1 vào biểu thức \(A=6x^3y-7xy\), ta được:

\(6\cdot1^3\cdot\left(-1\right)-7\cdot1\cdot\left(-1\right)\)

\(=-6+7=1\)

Vậy: 1 là giá trị của biểu thức \(A=6x^3y-7xy\) tại x=1 và y=-1

Chọn D

e) \(8\left(x+3y\right)-16x\left(x+3y\right)=\left(x+3y\right)\left(8-16x\right)=8\left(x+3y\right)\left(1-2x\right)\)

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

g) \(3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)=\left(3+5x\right)\left(x-y\right)\)

mn ơiiiiiiiiiiiii

a,b) \(n\ge 3\)

Câu 2: C