= (x+1-y)(x+1+y)

hằng đẳng thức số 3: a^2 - b^2 = (a-b)(a+b)

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

( x + 1 )² - y²

= ( x - y + 1 )( x + y + 1 )

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

= ( x-y).(x+y)

\(\left(3x^2+\dfrac{2}{3}y\right)^2\)

\(=\left(3x^2\right)^2+2.3x.\dfrac{2}{3}y+\left(\dfrac{2}{3}y\right)^2\)

\(=9x^4+4xy+\dfrac{4}{9}y^2\)

(3x²+\(\dfrac{2}{3}\)y)²

=9\(x^4\)+\(\dfrac{4}{9}\)\(y^2\)

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

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

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

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

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

(2/3x+3y).(3y-2/3x)

= ( 3y)^2  - ( 2/3 x )^2

=  9y^2  - 4/9 x^2

`(2/3 x+3y)(3y-2/3x)`

`=(3y +2/3x)(3y-2/3x)`

`= (3y)^2 - (2/3x)^2`

`= 9y^2 - 4/9 x^2`

Vận dụng : `A^2-B^2=(A-B)(A+B)`

Ta có: \(x^2\cdot\left(x^4+25\right)\cdot\left(x^2-5\right)\cdot\left(x^2+5\right)\cdot\left(x-y\right)\left(x^2+xy+y^2\right)\cdot\left(x^3+y^3\right)\)

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

\(=x^2\cdot\left(x^8-625\right)\cdot\left(x^6-y^6\right)\)

a) \(=x^3+27-54-x^3=-27\)

b) \(=8x^3+y^3\)

a, (x+y)² = x² + 2xy + y²
b, ( x-4y)²= x² -8xy² + 16y2
c, \(\left(3x+\frac{1}{3}\right)^2=9x^2+2xy+\frac{1}{9}\)
d,\(4x^2-81=\left(2x-9\right)\left(2x+9\right)\)
e,\(\left(xy+5\right)^2=x^2y^2+10xy+25\)
f,\(\left(x-y+z\right)^2=x^2+y^2+z^2-2xy+2xz-2yz\)
g,\(1-9y^2=\left(1-3y\right)\left(1+3y\right)\)
h,\(\left(m-\frac{2}{3}n\right)^2=m^2-\frac{4}{3}mn+\frac{4}{9}n^2\)

(2x-1)(2x+1)(4x² +1)

= (4x² - 1)(4x²+1)

=16x⁴ - 1 

(2x-1).(2x+1).(4x^2+1)

= (4x^2 - 1).(4x^2+1)

=  (4x^2)^2  -  1

=  16x^4  - 1

`(2x-1)(2x+1)(4x^2+1)`

`=[(2x)^2-1^2](4x^2+1)`

`=(4x^2-1)(4x^2+1)`

`=(4x^2)^2-1^2`

`=16x^4-1`