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

2x³ + 3x² - 2x

= x ( 2x² + 3x - 2 )

= x ( 2\(x^2\) + 4\(x-x-2\) )

= x [ ( 2\(x^2\) + 4x ) - ( x + 2 )]

= x  [  2x ( x + 2 ) - ( x + 2 )]

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

Bài 1: 

a: \(8x^3-2x=2x\left(4x^2-1\right)=2x\left(2x-1\right)\left(2x+1\right)\)

c: \(-5m^3\left(m+1\right)+m+1=\left(m+1\right)\left(-5m^3+1\right)\)

Bài 1: 

a: \(=6x^3-10x^2+6x\)

b: \(=-2x^3-10x^2-6x\)

Bài 4: 

a: =>3x+10-2x=0

=>x=-10

c: =>3x²-3x²+6x=36

=>6x=36

hay x=6

Bài 1:

\(a,=6x^3-10x^2+6x\\ b,=-2x^3-10x^2-6x\)

Bài 4:

\(a,\Leftrightarrow3x+10-2x=0\Leftrightarrow x=-10\\ b,\Leftrightarrow x\left(2x^2+9x-5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\\ \Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4,5=3,5\\ \Leftrightarrow-6x=8\Leftrightarrow x=-\dfrac{4}{3}\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\)

Bài 1:

\(a,=7xy\left(2x-3y+4xy\right)\\ b,=x\left(x+y\right)-5\left(x+y\right)=\left(x-5\right)\left(x+y\right)\\ c,=\left(x-y\right)\left(10x+8\right)=2\left(5x+4\right)\left(x-y\right)\\ d,=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\\ =2x\left(4x+2\right)=4x\left(2x+1\right)\\ e,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x^2+8x-x-8=\left(x+8\right)\left(x-1\right)\\ g,\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\\ h,=x^2+3x+x+3=\left(x+3\right)\left(x+1\right)\)

a: Ta có: \(x^2-4y^2-2x-4y\)

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

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

c: Ta có: \(x^3+2x^2y-x-2y\)

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

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

d: Ta có: \(3x^2-3y^2-2\cdot\left(x-y\right)^2\)

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

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

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

e: Ta có: \(x^3-4x^2-9x+36\)

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

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

f: Ta có: \(x^2-y^2-2x-2y\)

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

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

Bài 3: 

b: \(x^2+2x+1=\left(x+1\right)^2\)

c: \(x^2-16=\left(x-4\right)\left(x+4\right)\)

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

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

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

1/(x+2)² -(3x-1)²=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x²+12x

2/(x⁴+x²)(-2x³-2x)=x²(x²+1)-2x(x²+1)=(x²+1)(x²-2x)

Bài 1: 

b: \(3x-6=x^2-16\)

\(\Leftrightarrow x^2-3x-10=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

Câu 6:Thực hiện phép nhân  -2x(x² + 3x - 4) ta được:

A.-2x³ - 6x² – 8x          B. 2x³ -6x² – 8x      C. -2x³ - 6x² + 8x         D. -2x³ + 3x² -4

Câu 7 : Phân tích đa thức x² + 2xy + y² – 9z² thành nhân tử ta được:

A. (x+y+3z)(x+y–3z)  

B. (x-y+3z)(x+y–3z) 

C.(x - y +3z)(x - y – 3z)

D. (x + y +3z)(x -y – 3z)

Câu 9: Phân tích đa thức x² + 7x + 12 thành nhân tử ta được:

A. (x - 3)( x + 4 )         B. (x + 3)( x + 4 )         C.(x + 5)( x + 2 )               D. (x -5)( x + 2 )

Câu 10:  Giá trị của biểu thức  (x² + 4x + 4) tại x = - 2 là:

A. 4                            B. -2                          C. 0                           D. -8

Mấy câu còn lại bị lỗi r nhé

Câu 6:C

Câu 7:A

Câu 9:B

Câu 10:A

x⁴ - 2x³ + 2x - 1

= (x⁴ - 1) - (2x³ - 2x)

= (x² - 1)(x² + 1) - 2x(x² - 1)

= (x² - 1)(x² + 1 - 2x)

= (x - 1)(x + 1)(x² - 2x + 1)

= (x - 1)(x + 1)(x - 1)²

= (x - 1)³(x + 1)