`1)(a^[1/4]-b^[1/4])(a^[1/4]+b^[1/4])(a^[1/2]+b^[1/2])`

`=[(a^[1/4])^2-(b^[1/4])^2](a^[1/2]+b^[1/2])`

`=(a^[1/2]-b^[1/2])(a^[1/2]+b^[1/2])`

`=a-b`

`2)(a^[1/3]-b^[2/3])(a^[2/3]+a^[1/3]b^[2/3]+b^[4/3])`

`=(a^[1/3]-b^[2/3])[(a^[1/3])^2+a^[1/3]b^[2/3]+(b^[2/3])^2]`

`=(a^[1/3])^3-(b^[2/3])^3`

`=a-b^2`

(1𝑦/3+3)^3

(𝑦/3+3)^3

(𝑦/3+3⋅3/3)^3

(𝑦+3⋅3/3)^3

(𝑦+9/3)^3

\(\left(\dfrac{1}{3}y+3\right)^3=\dfrac{1}{27}y^3+y^2+9y+27\)

\(\left(\dfrac{1}{3y+3}\right)^3=\dfrac{1}{\left(3y+3\right)^3}=\dfrac{1}{27y^3+81y^2+81y+27}\)

\(\left(\dfrac{1}{3y+3}\right)^3=\dfrac{1^3}{\left(3y+3\right)^3}=\dfrac{1}{27\left(y^3+3y^2+3y+1\right)}\)

a: Ta có: \(\left(x+y\right)^3-\left(x-y\right)^3\)

\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3\)

\(=6x^2y+2y^3\)

\(\left(x+y\right)^3-\left(x-y\right)^3\)

\(=\left(x+y-x+y\right)^3+3\left(x+y\right)\left(x-y\right)\left(x+y-x+y\right)\)

\(=8y^3+6y\left(x^2-y^2\right)\)

\(=8y^3+6x^2y-6y^3\)

\(=2y^3+6x^2y\)

a: \(\left(6x^2+\frac13\right)^2\)

\(=\left(6x^2\right)^2+2\cdot6x^2\cdot\frac13+\left(\frac13\right)^2\)

\(=36x^4+4x^2+\frac19\)

b: \(\left(5x-4y\right)^2=\left(5x\right)^2-2\cdot5x\cdot4y+\left(4y\right)^2\)

\(=25x^2-40xy+16y^2\)

\(\left(x+y\right)^3=x^3+3x^2y+3xy^2-y^3\)

\(\left(x-y\right)^3=x^3-3x^2y+3xy^2-y^3\)

\(\left(2y-3\right)^3=8y^3-36y^2+54y-27\)

a: Ta có: \(\left(x+y\right)^3-\left(x-y\right)^3\)

\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3\)

\(=6x^2y+2y^3\)

Lời giải:

\(A=(x+2y+3z)(x-2y+3z)\)

\(=[(x+3z)+2y][(x+3z)-2y]\)

\(=(x+3z)^2-(2y)^2\)

\(=x^2+9z^2+6xz-4y^2\)

Ta có:(a²+ab+b²)(a²-ab+b²)-(a⁴+b⁴)

    =   (a²+b²)²-a²b²-a⁴-b⁴=a⁴+2a²b²+b⁴-a²b²-a⁴-b⁴=a²b²

Ta có:(a²+ab+b²)(a²-ab+b²)-(a⁴+b⁴)

    =   (a²+b²)²-a²b²-a⁴-b⁴=a⁴+2a²b²+b⁴-a²b²-a⁴-b⁴=a²b²

Dùng thẳng thì nhanh hơn \(\left(a+b\right)^2-\left(a-b\right)^2=\left(a+b-a+b\right)\left(a+b+a-b\right)\)

\(\left(a+b\right)^2-\left(a-b\right)^2\)

\(=\left(a^2+2ab+b^2\right)-\left(a^2-2ab+b^2\right)\)

\(=a^2+2ab+b^2-a^2+2ab-b^2\)

\(=4ab\)

a) $(3x+5)^2\\=(3x)^2+2.3x.5+5^2\\=9x^2+30x+25$

b) $(6x+\dfrac{1}{3})^2\\=(6x)^2+2.6x.\dfrac{1}{3}+(\dfrac{1}{3})^2\\=36x^2+4x+\dfrac{1}{9}$

c) $(5x-4y)^2\\=(5x)^2-2.5x.4y+(4y)^2\\=25x^2-40xy+16y^2$

d) $(5x-3)(5x+3)\\=(5x)^2-(3)^2\\=25x^2-9$