Ta có : 

a.  $(2 + xy) ^{2}$

=  $22 + 2 . 2 . xy + (xy) ^{2}$

=  $4 + 4xy + x ^{2}y ^{2}$

Vậy $(2 + xy) ^{2}$ = $4 + 4xy + x ^{2}y ^{2}$

b.  $(5 – 3x) ^{2}$

=  $5 ^{2}– 2 . 5 . 3x + (3x) ^{2}$

=  $25 – 30x + 9x ^{2}$

Vậy $(5 – 3x) ^{2}$ = $25 – 30x + 9x ^{2}$

c.  $(5 – x^{2} )(5 + x^{2} )$

=  $5 ^{2}– (x^{2} ) ^{2}$

=  $25 – x ^{4}$

Vậy $(5 – x^{2} )(5 + x^{2} )$ = $25 – x ^{4}$

d.  $(5x – 1) ^{3}$

=  $(5x) ^{3}– 3 . (5x)^{2} . 1 + 3 . 5x . 1 ^{2}– 1 ^{3}$

=  $125x ^{3}– 75x ^{2}+ 15x – 1$

Vậy  $(5x – 1) ^{3}$ = $125x ^{3}– 75x ^{2}+ 15x – 1$

e.  $(2x – y)(4x ^{2}+ 2xy + y ^{2})$

=   $(2x – y)[(2x) ^{2}+ 2x . y + y ^{2}] $

=   $(2x) ^{3}– y ^{3}= 8x ^{3}– y ^{3}$

Vậy $(2x – y)(4x ^{2}+ 2xy + y ^{2})$ = $8x ^{3}– y ^{3}$

f.  $(x + 3)(x ^{2}– 3x + 9)$

=  $(x + 3)(x ^{2}– 3x + 3^{2} )$

=   $x ^{3}+ 3 ^{3}= x ^{3}+ 27.$

Vậy $(x + 3)(x ^{2}– 3x + 9)$ = $x ^{3}+ 27.$