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

\(=x^2\left(x-1\right)-2x\cdot\left(x-1\right)\)

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

5: \(y^2\left(x^2+y\right)-z\cdot x^2-zy\)

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

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

7: \(5\left(x+y\right)^2+15\left(x+y\right)\)

\(=5\left(x+y\right)\cdot\left(x+y\right)+5\left(x+y\right)\cdot3\)

=5(x+y)(x+y+3)

9: \(7x\left(y-4\right)^2-\left(4-y\right)^3\)

\(=7x\left(y-4\right)^2+\left(y-4\right)^3\)

\(=\left(y-4\right)^2\left(7x+y-4\right)\)

11: \(\left(x+1\right)\left(y-2\right)-\left(2-y\right)^2\)

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

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

=(y-2)(x+y+3)

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

\(=\left(x-2\right)\left(5x-3x^2\right)=x\left(5-3x\right)\left(x-2\right)\)

4: 3x(x-5y)-2y(5y-x)

=3x(x-5y)+2y(x-5y)

=(x-5y)(3x+2y)

6: b(a-c)+5c-5a

=b(a-c)-5(a-c)

=(a-c)(b-5)

8: 9x(x-y)-10(y-x)^2

=9x(x-y)-10(x-y)^2

=(x-y)(9x-10x+10y)

=(x-y)(-x+10y)

10: \(\left(a-b\right)^2-\left(a+b\right)\left(b-a\right)\)

\(=\left(a-b\right)^2+\left(a+b\right)\left(a-b\right)\)

=(a-b)(a-b+a+b)

=2a(a-b)

12: 2x(x-3)+y(x-3)+(3-x)

=(x-3)(2x+y)-(x-3)

=(x-3)(2x+y-1)

a)x⁷+x⁵+1=x⁷+x⁶-x⁶+2x⁵-x⁵+x⁴-x⁴+x³-x³+x²-x²+1

=x⁷-x⁶+x⁵-x³+x²+x⁶-x⁵+x⁴-x²+x+x⁵-x⁴+x³-x+1

=x²(x⁵-x⁴+x³-x+1)+x(x⁵-x⁴+x³-x+1)+1(x⁵-x⁴+x³-x+1)

=(x²+x+1)(x⁵-x⁴+x³-x+1)

b)4x⁴-32x²+1=4x⁴+12x³+2x²-12x³-36x²-6x+2x²+6x+1

=2x²(2x²+6x+1)-6x(2x²+6x+1)+1(2x²+6x+1)

=(2x²-6x+1)(2x²+6x+1)

c)x⁶+27=(x²+3)(x²-3x+3)(x²+3x+3)

d)3(x⁴+x²+1)-(x²+x+1)

=3x⁴-3x³+2x²+3x³-3x²+2x+3x²-3x+2

=x²(3x²-3x+2)+x(3x²-3x+2)+1(3x²-3x+2)

=(x²+x+1)(3x²-3x+2)

e)bạn tự làm nhé

2: \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

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

Đặt \(x^2+x+1=a\)ta có

\(a\left(a+1\right)-12=a^2+a-12=a^2+4a-3a-12=a\left(a+4\right)-3\left(a+4\right)=\left(a+4\right)\left(a-3\right)\)

Thay \(a=x^2+x+1\)ta được

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

\(=\left(x^2+x+5\right)\left(x^2+2x-x-2\right)=\left(x^2+x+5\right)\left[x\left(x+2\right)-\left(x+2\right)\right]=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)Kl...

3. \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

\(=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)

\(=\left(x^2+8x+7\right)\left(x^2+8x+7+8\right)+15\)

Đặt \(x^2+8x+7=a\) Ta có

\(a\left(a+8\right)+15=a^2+8a+15=a^2+5a+3a+15=a\left(a+5\right)+3\left(a+5\right)=\left(a+5\right)\left(a+3\right)\)

Thay \(a=x^2+8x+15\)ta được

\(\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)

\(=\left(x^2+6x+2x+12\right)\left(x^2+8x+10\right)\)

\(=\left(x+6\right)\left(x+2\right)\left(x^2+8x+10\right)\)

1) Ta có : 2x² + 3x - 5

= 2x² - 2x + 5x - 5

= 2x(x - 1) + 5(x - 1)

= (x - 1) (2x + 5) 

3) x² + x - 6

= x² + 2x - 3x - 6

= x(x + 2) - (3x + 6)

= x(x + 2) - 3(x + 2)

= (x - 3)(x + 2) 

M = x⁹ - x⁷ + x⁶ - x⁵ - x⁴ + x³ - x² + 1

= ( x⁹ - x⁷ ) + ( x⁶ - x⁴ ) - ( x⁵ - x³ ) - ( x² - 1 )

= x⁷( x² - 1 ) + x⁴( x² - 1 ) - x³( x² - 1 ) - ( x² - 1 )

= ( x² - 1 )( x⁷ + x⁴ - x³ - 1 )

= ( x - 1 )( x + 1 )[ x⁴( x³ + 1 ) - ( x³ + 1 ) ]

= ( x - 1 )( x + 1 )( x³ + 1 )( x⁴ - 1 )

= ( x - 1 )( x + 1 )( x + 1 )( x² - x + 1 )( x² - 1 )( x² + 1 )

= ( x + 1 )²( x - 1 )( x² - x + 1 )( x - 1 )( x + 1 )( x² + 1 )

= ( x + 1 )³( x - 1 )²( x² + 1 )( x² - x + 1 )

7,x²-2xy+y²+3x-3y=(x-y)²+3(x-y)=(x-y)(x-y+3)

8,x⁴+4=(x⁴+4x²+4)-4x²=(x²+2)²-(2x)²=(x²-2x+2)(x²+2x+2)

9,4x(x+1)²-5x²(x+1)-4.(x+1)=(x+1)\(\left[4x\left(x+1\right)-5x^2-4\right]\)=(x+1)(4x²+4x-5x²-4)=(x+1)(-x²+4x-4)=-(x+1)(x-2)²