1. 2x(3x² - 5x + 3) = 6x³ - 10x² + 6x

2. \(-\dfrac{1}{2}x^2\left(2x^3-4x+3\right)=-x^5+2x^3+\dfrac{-3}{2}x^2\)

3. -2x(x² + 5x - 3) = -2x³ - 10x² + 6x

4. x(3x² - 2x + 5) = 3x³ - 2x² + 5x

5. 3xy²(2x - 4y + 3xy) = 6x²y² - 12xy³ = 9x²y³

Bài 6:

c: \(9x^2+6x+1=\left(3x+1\right)^2\)

d: \(4x^2-9=\left(2x-3\right)\left(2x+3\right)\)

e: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)

\(A=\dfrac{3}{2}x^2+6x\)

bậc 2 

\(\frac{1}{2}x^2y^2\left(2x+y\right)\left(2x-y\right)=\frac{1}{2}x^2y^2\left(4x^2-y^2\right)=2x^4y^2-\frac{1}{2}x^2y^4\)

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

Đa thức này không phân tích được bạn ơi

Q= (x² - 2y² + 3/4xy) - (2x² - y² + 3/4xy)

Q = x² - 2y² + 3/4xy - 2x² + y² - 3/4xy

Q= (x² - 2x²) + (-2y² + y²) + (3/4xy - 3/4xy)

Q= -x² - y² 

#Hk_tốt

#Ken'z

\(\left(2x^2-y^2+\frac{3}{4}xy\right)+Q=x^2-2y^2+\frac{3}{4}xy\)

\(\Rightarrow Q=x^2-2y^2+\frac{3}{4}xy-2x^2+y^2-\frac{3}{4}xy\)

\(\Rightarrow Q=-x^2-y^2\)

Vậy \(Q=-x^2-y^2\)

(x-1)(x+1)(x+2)

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

=x^3+2x^2-x-2

[X-1/2] [X+1/2] [4X-1]

=\(\left(x^2-\frac{1}{4}\right)\left(4x-1\right)\)

=\(4x^3-x^2-x+\frac{1}{4}\)

1/2X²Y² [2X+Y] [2X-Y]

=\(\frac{1}{2}x^2y^2\left(4x^2-y^2\right)\)

=\(2x^2y^2-\frac{1}{2}x^2y^4\)

a. 2x^2 ( 5x^3 - 4x^2 - 7xy + 1 )

= 10x^5 - 8x^4 - 14x^3y + 1

b. ( x - 5 ) ( x + 3 )

= x^2 + 3x - 5x - 15

= x^2 - 2x - 15

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

= x^2 + 2x - x - 2

= x^2 - x - 2

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

= 4x^2 - 2xy + 2xy - y^2

= 4x^2 - y^2

a) \(2x^2\left(5x^3-4x^2.g-7xy+1\right)\)

\(=10x^5-8x^4.g-14x^3y+2x^2\)

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

\(=x^2+3x-5x-15\)

\(=x^2-2x-15\)

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

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

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

d) \(\left(2x+y\right)\left(2x-y\right)\)

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

a) 2x² ( 5x³ - 4x² . g - 7xy + 1)
= 10x⁵ - 8x⁴g - 14x³y + 2x^2
b) (x-5)(x+3)
= x² + 3x - 5x - 15
= x² - 2x -15

 

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