a, 2xy^2 ( x^3 -3xy - 4 )

b, x^2 - 4x - 4x +16

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

= (x-4) (x-4)

Lời giải:

a.

$2x^4y^2-6x^2y^3-8xy^2=2xy^2(x^3-3xy-4)$

b. 

$x^2-8x+16=x^2-2.4.x+4^2=(x-4)^2$

c.

$12x^2-12=12(x^2-1)=12(x-1)(x+1)$

d. 

$5x^2y-20xy+20y=5y(x^2-4x+4)=5y(x-2)^2$
e.

$3x^2y^2-27y^2=3y^2(x^2-9)=3y^2(x-3)(x+3)$

f. 

$8x^3-27y^3=(2x)^3-(3y)^3=(2x-3y)(4x^2+6xy+9y^2)$

g.

$4x^4-8x^3+4x^2=(2x^2)^2-2.2x^2.2x+(2x)^2$

$=(2x^2-2x)^2=[2x(x-1)]^2=4x^2(x-1)^2$
h.

$7x^2y^2-28y^4=7y^2(x^2-4y^2)=7y^2(x-2y)(x+2y)$

$

\(4x^2-5xy+y^2=\left(4x^2-5xy+\dfrac{25}{16}y^2\right)-\dfrac{9}{16}y^2=\left(2x-\dfrac{5}{4}y\right)^2-\dfrac{9}{16}y^2=\left(2x-\dfrac{5}{4}y-\dfrac{3}{4}y\right)\left(2x-\dfrac{5}{4}y+\dfrac{3}{4}y\right)=\left(2x-2y\right)\left(2x-\dfrac{1}{2}y\right)=\left(x-y\right)\left(4x-y\right)\)

\(4x^2-5xy+y^2\)

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

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

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

a) \(4x^2-9y^2+6x-9y\)

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

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

b) \(1-2x+2yz+x^2-y^2-z^2\)

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

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

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

Tick hộ mình nha 😘

\(A=x^2-y^2+7x+7y\)

\(=\left(x-y\right)\left(x+y\right)+7\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y+7\right)\)

\(B=4x^3-4x^2+x\)

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

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

\(C=x^2-6xy+9y^2-9\)

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

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

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

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

C=\(x^2+9y^2-9-6xy=\left(x^2-6xy+9y^2\right)-9=\left(x-3y\right)^2-3^2=\left(x-3y-3\right)\left(x-3y+3\right)\)

a: =3x(y-2)-5(y-2)

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

b: =(2x-5y)(2x+5y)

c: =x(x^3-8)

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

a)

\(3x\left(y-2\right)+5\left(2-y\right)\\=3x\left(y-2\right)-5\left(y-2\right)\\ =\left(y-2\right)\left(3x-5\right) \)

b)

\(4x^2-25y^2\\ =\left(2x-5y\right)\left(2x+5y\right)\)

c)

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

\(a,3x\left(y-2\right)+5\left(2-y\right)=3x\left(y-2\right)-5\left(y-2\right)=\left(y-2\right)\left(3x-5\right)\)

\(b,4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)

\(c,x^4-8x=x\left(x^3-8\right)=x\left(x-2\right)\left(x^2+4x+4\right)\)

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

b) \(\left(x^2+x\right)^2+4x^2+4x-12=\left(x^2+x\right)^2+4\left(x^2+x\right)+4-16=\left(x^2+x+2\right)^2-4^2=\left(x^2+x+2-4\right)\left(x^2+x+2+4\right)=\left(x^2+x-2\right)\left(x^2+x+6\right)=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)

c) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)+1-25=\left(x^2+7x+11\right)^2-5^2=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

a. \(x^2\left(x^2+4\right)-x^2-4\)

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

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

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

b. \(\left(x^2+x\right)^2+4x^2+4x-12\)

\(=x^4+2x^3+5x^2+4x-12\)

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

c. \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\) (*)

Đặt \(t=x^2+7x+10\), ta được

(*) \(=t\left(t+2\right)-24\)

\(=t^2+2t-24\)

\(=\left(t-4\right)\left(t+6\right)\)

hay \(\left(x^2+7x+6\right)\left(x^2+7x+18\right)\)

a: Ta có: \(x^2\left(x^2+4\right)-x^2-4\)

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

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

b: Ta có: \(\left(x^2+x\right)^2+4x^2+4x-12\)

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

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

\(=\left(x^2+x-2\right)\left(x^2+x+6\right)\)

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

c: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)

a: \(3abc^3-6a^2b^3c+12a^3bc\)

\(=3abc\cdot c^2-3abc\cdot2ab^2+3abc\cdot4a^2\)

\(=3abc\left(c^2-2ab^2+4a^2\right)\)

b: \(27-8y^3\)

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

\(=\left(3-2y\right)\left(9+6y+4y^2\right)\)

c: Sửa đề: \(4x^2+4x-y^2+1\)

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

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

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

d: \(3a^2\cdot\left(x-2\right)-6ab\cdot\left(2-x\right)\)

\(=3a^2\cdot\left(x-2\right)+6ab\cdot\left(x-2\right)\)

\(=\left(x-2\right)\left(3a^2+6ab\right)\)

\(=3a\left(a+2b\right)\left(x-2\right)\)

a: (x^2+x)^2+4x^2+4x-12

=(x^2+x)^2+4(x^2+x)-12

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

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

b: =(x^2+8x)^2+22(x^2+8x)+105+15

=(x^2+8x)^2+22(x^2+8x)+120

=(x^2+8x+10)(x^2+8x+12)

=(x^2+8x+10)(x+2)(x+6)

c: =8x^2+12x-2x-3

=(2x+3)(4x-1)