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

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

c) \(2x^3-x^2-8x+4\)

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

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

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

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

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

e) \(2x^2-5x+2\)

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

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

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

Ta có: \(x\left(y^2-z^2\right)+y\left(z^2-x^2\right)+z\left(x^2-y^2\right)\)

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

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

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

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

\(=\left(y-z\right)\left[\left(zx-yz\right)-\left(x^2-xy\right)\right]\)

\(=\left(y-z\right)\left[z\left(x-y\right)-x\left(x-y\right)\right]\)

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

Ta có: \(4x^4+81\) đề phải như thế này chứ?!

\(=\left(4x^4+36x^2+81\right)-36x^2\)

\(=\left(2x^2+9\right)^2-\left(6x\right)^2\)

\(=\left(2x^2-6x+9\right)\left(2x^2+6x+9\right)\)

Ta có :

\(n^4+7\left(7+2n^2\right)\)

\(=n^4+49+14n^2\)

\(=\left(n^2+7\right)^2\)

Vì n là số nguyên lẻ nên n có dạng 2k + 1 với k là số nguyên 

 \(\Rightarrow\left(n^2+7\right)^2=\left[\left(2k+1\right)^2+7\right]^2\)

\(=\left[\left(4k^2+4k+1\right)+7\right]^2\)

\(=\left(4k^2+4k+8\right)^2\)

\(=\left[4k\left(k+1\right)+8\right]^2\)

Vì \(\hept{\begin{cases}k\left(k+1\right)⋮2\forall k\in Z\\4⋮4\end{cases}}\) nên \(4k\left(k+1\right)⋮8\forall k\in Z\)

\(\Rightarrow4k\left(k+1\right)+8⋮8\forall k\in Z\)

\(\Rightarrow\left[4k\left(k+1\right)+8\right]^2⋮8^2\forall k\in Z\)

\(\Rightarrow\left[4k\left(k+1\right)+8\right]⋮64\forall k\in Z\)

=> đpcm 

n⁴ + 7( 7 + 2n² )

= n⁴ + 14n² + 49

= ( n² + 7 )²

Vì n lẻ và n ∈ Z => n = 2k + 1 ( k ∈ Z )

Thế vô ta được :

[ ( 2k + 1 )² + 7 ]²

= ( 4k² + 4k + 1 + 7 )²

= ( 4k² + 4k + 8 )²

= [ 4( k² + k + 2 ) ]²

= { 4[ k( k + 1 ) + 2 ] }²

Ta có : k( k + 1 ) chia hết cho 2

            2 chia hết cho 2

=> k( k + 1 ) + 2 chia hết cho 2

=> 4[ k( k + 1 ) + 2 ] chia hết cho 8

=>  { 4[ k( k + 1 ) + 2 ] }² chia hết cho 64

=> đpcm