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

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

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

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

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

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)

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

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

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

c) \(x^4-4x^3+8x^2-16x+16=x^4+8x^2+16-\left(4x^3+16x\right)\)

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

b) \(x^6-x^4-9x^3+9x^2=x^4\left(x^2-1\right)-\left(9x^3-9x^2\right)\)

\(=x^4\left(x-1\right)\left(x+1\right)-9x^2\left(x-1\right)\)

\(=\left(x^5+x^4-9x^2\right)\left(x-1\right)=\left(x-1\right)x^2\left(x^3+x^2-9\right)\)

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

a,x⁴-4x³+8x²-16x+16

=(x⁴-4x³+4x²⁾+(4x²-16x+16)

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

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

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

\(4x^4-16-4x^2-16x\)

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

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

\(=\left(x+1\right)\left[4x^2\left(x-1\right)-16\right]\)

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

Nguyễn Văn Tuấn AnhNs r, không biết thì not làm

\(4x^4-16-4x^2-16x\)

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

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

\(=\left(x+1\right)\left[4x^2\left(x-1\right)-16\right]\)

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

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

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

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

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

๖ۣۜMαɾαƙαĭ ⁹⁀ᶦᵈᵒᶫ  bạn đừng ns thế người ta không biết làm nhưng vẫn cố gắng giúp 

nếu bạn ấy làm sai thì vẫn có những người như bạn sửa

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

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

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

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

b )=x⁴-2x³-2x³+4x²+4x²-8x-8x+16

=x³(x-2)-2x²(x-2)+4x(x-2)-8(x-2)

=(x-2)(x³-2x²+4x-8)

=(x-2)[x²(x-2)+4(x-2)]

=(x-2)²(x²+4)

a) đề thiếu ko bn?

b) \(x^4-4x^3+8x^2-16x+16=\left(x^4-4x^2\right)-\left(4x^3-12x^2+8x\right)-\left(8x-16\right)\)

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

\(=x^2\left(x-2\right)\left(x+2\right)-4x\left(x-2\right)\left(x-1\right)-8\left(x-2\right)\)

\(=\left(x-2\right)\left[x^2\left(x+2\right)-4x\left(x-1\right)-8\right]=\left(x-2\right)\left(x^3-2x^2+4x-8\right)\)

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

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