\(=\left(x-3\right)\left(8x^3-16x^2\right)=8x^2\left(x-2\right)\left(x-3\right)\)

\(8x^3\left(x-3\right)+16x^2\left(3-x\right)\)

\(=8x^3\left(x-3\right)-16x^2\left(x-3\right)\)

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

\(x^4-4x^3+8x^2-16x+16 \)

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

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

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

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

a,x⁴-4x³+8x²-16x+16

=x⁴-4x³+4x²+4x²-16x+16

=x².(x-2)²+4.(x-2)²

=(x-2)²(x²+4)

Lời giải:

a. 

$(xy)^2-xy-2=(x^2y^2+xy)-(2xy+2)$

$=xy(xy+1)-2(xy+1)=(xy+1)(xy-2)$

b. Bạn xem lại đoạn $-16x^2$ là dấu - hay + vậy?

x^4 - 4x^3 - 8x^2 - 16x + 16 
= x^4-8x^2+16-4x^3-16x 
= ( x^2+4)^2 - 4x(x^2+4 ) 
= ( x^2 + 4 )(x^2 + 4 - 4x) 
= (x^2 + 4 )( x - 2 )^2

\(x^4-4x^3+8x^2-16x+16\)

=\(x^4-4x^3+4x^2-16x+16\)

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

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

x⁴ - 4x³ +  8x²  - 16x + 16 = x⁴ - 4x³ +  4x²+4x²  - 16x + 16= ( x⁴ - 4x³ +  4x²)+(4x²  - 16x + 16)

=x²(x²-4x+4) + 4( x²-4x+4) = (x²+4)(x²-4x+4)=(x²+4)(x-2)²

=(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)`

\(16x-5x^2-3=-\left(5x^2-16x+3\right)=-\left[\left(5x^2-15x\right)-\left(x-3\right)\right]=-\left[5x\left(x-3\right)-\left(x-3\right)\right]=-\left(5x-1\right)\left(x-3\right)=\left(5x-1\right)\left(3-x\right)\)