a)  \(27x^3-8=\left(3x-2\right)\left(9x^2+6x+4\right)\)

b)  \(8x^3+12x^2+6x+1=\left(2x+1\right)^3\)

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

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

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

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

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

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

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

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

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

.

hk tôt

\(a,x^2\left(x-2\right)-4x+8\\ =\left(x^2-4\right)\left(x-2\right)\\ =\left(x-2\right)^2\left(x+2\right)\\ b,x^2+7xy+10y^2\\ =x^2+2xy+5xy+10y^2\\ =x\left(x+2y\right)+5y\left(x+2y\right)\\ =\left(x+5y\right)\left(x+2y\right)\)

6x²+7xy+2y²

=6x²+3xy+4xy+2y²

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

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

phân tích đa thức thành nhân tử

a, 6x^2 + 7xy + 2y^2

=6x^2+3xy+4xy+2y^2

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

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

b, 9x^2 - 9xy - 4y^2

=9x^2 +3xy-12xy-4y^2

=3x(x+y)-4y(x+y)

=(3x+4y)(x+y)

c, x^2 - y^2 + 10x - 6y + 16=x^2-y^2+6x-6y+4x+16=x(x+6)-y(x+6)+4(x+6)=(x-y+4)(x+6)

Bài làm

a, 6x² + 7xy + 2y²

= 6x² + 3xy + 4xy + 2y² 

= ( 6x² + 3xy ) + ( 4xy + 2y² )

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

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

b, 9x² - 9xy - 4y² 

= 9x² - 12xy + 3xy - 4y² 

= ( 9x² - 12xy ) + ( 3xy - 4y² )

= 3x( 3x - 4y ) + y ( 3x - 4y )

= ( 3x + y )( 3x - 4y )

c, x² - y² + 10x - 6y + 16

 = x² - y² - 6x + 6y + 4x + 16

= x( x + 6 ) - y( x + 6 ) + 4( x + 6 )

= ( x - y + 4 )( x + 6 )

# Học tốt #

a) x³ ​- 4x² - 12x + 27

= \(\left(x^3+3x^2\right)-\left(7x^2+21x\right)+\left(9x+27\right)\)

= \(\left(x+3\right)\left(x^2-7x+9\right)\)

b) 9x² + 6x - 8 

=\(9x^2-6x+12x-8=3x\left(3x-2\right)+4\left(3x-2\right)\)

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

c) x² - 7xy + 10y²

^{=\(x^2-5xy-2xy+10y^2=x\left(x-5y\right)-2y\left(x-5y\right)\)}

^{=\(\left(x-5y\right)\left(x-2y\right)\)}

a) x³ ​- 4x² - 12x + 27

 =x³ + 3x² - 7x² - 21x + 9x + 27 

= x²(x+3) - 7x(x+3) + 9(x+3)  

= (x² - 7x + 9)(x + 3)

b) 9x² + 6x - 8 

= 9x² - 6x + 12x - 8

= 3x(3x - 2) +  4(3x - 2) 

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

c) x² - 7xy + 10y²

= x² - 5xy - 2xy + 10y²

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

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

d)  x⁸ + x⁷ + 1 

Ta thêm vào các số hạng x⁶, x⁵, x⁴, x³, x², x và cùng bớt đi các số hạng ấy ta có:

= x⁸ - x⁶ + x⁵ - x³ + x² + x⁷ - x⁵ + x⁴ -x² +x + x⁶ - x⁴ + x³ - x + 1

= x²(x⁶ - x⁴ + x³ - x + 1) + x(x⁶ - x⁴ + x³ - x + 1) + x⁶ - x⁴ + x³ - x + 1

= (x² + x + 1)(x⁶ - x⁴ + x³ - x + 1)

a) \(x^3-4x^2-12x+27\)

\(=\left(x^3+27\right)-\left(4x^2+12x\right)\)

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

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

\(=\left(x+3\right)\left(x^2-7x+9\right)\)

b) \(9x^2+6x-8\)

\(=9x^2+6x+1-9\)

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

\(=\left(3x+1-9\right)\left(3x+1+9\right)\)

\(=\left(3x+8\right)\left(3x+10\right)\)

c) \(x^2-7xy+10y^2\)

\(=x^2-2xy-5xy+10y^2\)

\(=x\left(x-2y\right)-5y\left(x-2y\right)\)

\(=\left(x-2y\right)\left(x-5\right)\)

a)  \(x^2+6x+8\)

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

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

\(\left(x-2\right)\left(x-4\right)\)

b) \(x^2-7xy+10y^2\)

\(=x^2-2xy-5xy+10y^2\)

\(=x\left(x-2y\right)-5y\left(x-2y\right)\)

\(=\left(x-2y\right)\left(x-5y\right)\)

a) x² - 6x + 8

= x² -2x - 4x +8

= x( x-2) -4( x-2)

= ( x-2)(x-4)

a/ \(x^2-6x+8=x^2-2x-4x+8\) \(8\) \(=x\left(x-2\right)-4\left(x-2\right)\) \(=\left(x-4\right)\left(x-2\right)\)

b/ \(x^2-7xy+10y^2\) \(=x^2-2xy-5xy+10y^2\) \(=x\left(x-2y\right)-5y\left(x-2y\right)\) \(=\left(x-5y\right)\left(x-2y\right)\)

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

b: =x^2-6xy+9y^2=(x-3y)^2

c: =5x(x^2-2xy+y^2)

=5x(x-y)^2

d: =(x-3)^2

e: =(2y-z)(4x+7y)

a)HĐT:(x^2+1-x)(x^2+1+x)

b)=x^2-2.x.3y+(3y)^2

c)=5x(x^2-2xy+y^2)

=5x(x-y)^2

d)x^2-2.3.x+3^2

=(x-3)^2

e)(2y-z)+7y(2y-z)

=(2y-z)(1+7y)

Lời giải:

a.

\(-16a^4b^6-24a^5b^5-9a^6b^4=-[(4a^2b^3)^2+2.(4a^2b^3).(3a^3b^2)+(3a^3b^2)^2]\)

\(=-(4a^2b^3+3a^3b^2)^2=-[a^2b^2(4b+3a)]^2\)

\(=-a^4b^4(3a+4b)^2\)

b.

$x^3-6x^2y+12xy^2-8x^3$

$=x^3-3.x^2.2y+3.x(2y)^2-(2y)^3=(x-2y)^3$

c.

$x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}$

$=x^3+3.x^2.\frac{1}{2}+3.x.\frac{1}{2^2}+(\frac{1}{2})^3$

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

a) Ta có: \(-16a^4b^6-24a^5b^5-9a^6b^4\)

\(=-a^4b^4\left(16b^2+24ab+9a^2\right)\)

\(=-a^4b^4\cdot\left(4b+3a\right)^2\)

b) Ta có: \(x^3-6x^2y+12xy^2-8y^3\)

\(=x^3-3\cdot x^2\cdot2y+3\cdot x\cdot\left(2y\right)^2-\left(2y\right)^3\)

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

c) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}\)

\(=x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3\)

\(=\left(x+\dfrac{1}{2}\right)^3\)