a) Ta có: \(1+6x-6x^2-x^3\)

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

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

b:Ta có: \(x^3-4x^2+8x-8\)

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

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

c: Ta có: \(x^3+2x^2+2x+1\)

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

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

d: Ta có: \(8x^3-12x^2+6x-1\)

\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3\)

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

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.

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

$

a: \(50x^5-8x^3\)

\(=2x^3\left(25x^2-4\right)\)

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

b: \(x^4-5x^2-4y^2+10y\)

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

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

c: \(36a^2+12a+1-b^2\)

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

\(=\left(6a+1-b\right)\left(6a+1+b\right)\)

d: \(x^3+y^3-xy^2-x^2y\)

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

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

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

e: Ta có: \(4x^2+4x-3\)

\(=4x^2+6x-2x-3\)

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

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

f: Ta có: \(9x^4+16x^2-4\)

\(=9x^4+18x^2-2x^2-4\)

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

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

g: Ta có: \(-6x^2+5xy+4y^2\)

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

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

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

h: Ta có: \(\left(x^2+4x\right)^2+8\left(x^2+4x\right)+15\)

\(=\left(x^2+4x\right)^2+3\left(x^2+4x\right)+5\left(x^2+4x\right)+15\)

\(=\left(x^2+4x+3\right)\cdot\left(x^2+4x+5\right)\)

\(=\left(x+1\right)\left(x+3\right)\left(x^2+4x+5\right)\)

\(=\left(x+4\right)^2-1=\left(x+4-1\right)\left(x+4+1\right)=\left(x+3\right)\left(x+5\right)\)

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

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

=(x^2+8x)^2+23(x^2+8x)+135

Cái này ko phân tích được nha bạn

\(\left(x^2+8x+8\right)\left(x^2+8x+15\right)+15\\ \Leftrightarrow\left(x^4+8x^3+15x^2+8x^3+64x^2+120x+8x^2+64x+120\right)+15\\ \Leftrightarrow x^4+16x^3+87x^2+184x+135\)

Gọi `A=(x^2+8x+8)(x^2+8x+15)+15`

Đặt `t=x^2+8x+11,5`

`=>A=(t-3,5)(t+3,5)+15=t^2-3,5^2+15=t^2-2,75=(t-sqrt(2,75))(t+sqrt(2,75))=(x^2+8x+11,5-(sqrt11)/2)(x^2+8x+11,5+(sqrt11)/2)=(x^2+8x+(23-\sqrt11)/2)(x^2+8x+(23+\sqrt11)/2)`

( x² + 8x + 7 ) ( x² + 8x + 15 ) + 15

Đặt x² + 8x + 7 = y ta có:

   y ( y + 8 ) + 15 

= y² + 8y + 15 

= ( y + 3 ) ( y + 5 )

= ( x² + 8x + 10 ) ( x² + 8x + 12 ) 

= ( x² + 8x + 10 ) ( x + 2 ) ( x + 6 )

Đặt x² + 8x + 7 = y ta có:

   y ( y + 8 ) + 15 

= y² + 8y + 15 

= ( y + 3 ) ( y + 5 )

= ( x² + 8x + 10 ) ( x² + 8x + 12 ) 

= ( x² + 8x + 10 ) ( x + 2 ) ( x + 6 )

đặt t = x² + 8x + 7

có : t ( t + 8 ) + 15 

= t² + 8t + 15

= t² + 5t + 3t + 15

= ( t² + 5t ) + ( 3t + 15 )

= t ( t + 5 ) + 3 ( t + 5 )

= ( t + 3 ) ( t + 5 )

mà t = x² + 8x + 7

⇒ ( x² + 8x + 10 ) ( x² + 8x + 12 )

a) \(x^4+8x+63\)

\(=x^4+4x^3+9x^2-4x^3-16x^2-36x+7x^2+28x+63\)

\(=x^2\left(x^2+4x+9\right)-4x\left(x^2+4x+9\right)+7\left(x^2+4x+9\right)\)

\(=\left(x^2+4x+9\right)\left(x^2-4x+7\right)\)

c) \(\left(x^2+2x+7\right)+\left(x^2-2x+4\right)\left(x^2+2x+3\right)\left(1\right)\)

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

\(\Rightarrow x^2+2x+4=\dfrac{x^3-8}{x-2}\)

\(\left(1\right)\Rightarrow\left[\left(\dfrac{x^3-8}{x-2}+3\right)\right]+\left(x^2-2x+4\right)\left[\left(\dfrac{x^3-8}{x-2}-1\right)\right]\)

\(=\left[\left(\dfrac{x^3-3x-14}{x-2}\right)\right]+\left(x^2-2x+4\right)\left[\left(\dfrac{x^3-2x-5}{x-2}\right)\right]\)

\(=\dfrac{1}{x-2}\left[x^3-3x-14+\left(x^2-2x+4\right)\left(x^3-2x-5\right)\right]\)

\(8x^3-\frac{1}{8}\)

\(=\left(2x\right)^3-\left(\frac{1}{2}\right)^3\)

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

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