`@` `\text {Ans}`

`\downarrow`

`(2y + 3x^2)^3`

`= (2y)^3 + 3. (2y)^2 . 3x^2 + 3. 2y . (3x^2)^2 + (3x^2)^3`

`= 8y^3 + 3. 4y^2 . 3x^2 + 6y . 9x^4 + 27x^6`

`= 8y^3 + 36x^2y^2 +54x^4y + 27x^6`

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CT:

`(A+B)^3 = A^3 + 3A^2B + 3AB^2 + B^3`

Để triển khai biểu thức (2y + 3x^2)^3 bằng hằng đẳng thức, ta sử dụng công thức nhị thức Newton:

(2y + 3x^2)^3 = C(3, 0)(2y)^3(3x^2)^0 + C(3, 1)(2y)^2(3x^2)^1 + C(3, 2)(2y)^1(3x^2)^2 + C(3, 3)(2y)^0(3x^2)^3

Trong đó:
C(n, k) là tổ hợp chập k của n (C(n, k) = n! / (k!(n-k)!))
^ là dấu mũ
() là dấu ngoặc

Áp dụng công thức, ta có:

(2y + 3x^2)^3 = C(3, 0)(2y)^3(3x^2)^0 + C(3, 1)(2y)^2(3x^2)^1 + C(3, 2)(2y)^1(3x^2)^2 + C(3, 3)(2y)^0(3x^2)^3
= 1(2y)^3 + 3(2y)^2(3x^2) + 3(2y)(3x^2)^2 + 1(3x^2)^3
= 8y^3 + 12y^2(3x^2) + 6y(9x^4) + 27x^6
= 8y^3 + 36y^2x^2 + 54yx^4 + 27x^6

Vậy biểu thức (2y + 3x^2)^3 sau khi triển khai bằng hằng đẳng thức là 8y^3 + 36y^2x^2 + 54yx^4 + 27x^6.

\(a.\left(2xy-3\right)^2=4x^2y^2-12xy+9\)

\(b.\left(\dfrac{1}{2}x+\dfrac{1}{3}\right)^2=\dfrac{1}{4}x^2+\dfrac{1}{3}x+\dfrac{1}{9}\)

a) (2xy)²-2(2xy-3)+3²

a)\(\left(2xy-3\right)^2=\left(2xy\right)^2-2\cdot2xy\cdot3+3^2=4x^2y^2-12xy+9\)

b)\(\left(\dfrac{1}{2}x+\dfrac{1}{3}y\right)^2=\left(\dfrac{1}{2}x\right)^2+2\cdot\dfrac{1}{2}x\cdot\dfrac{1}{3}y+\left(\dfrac{1}{3}y\right)^2\)

                          \(=\dfrac{1}{4}x^2+\dfrac{1}{3}xy+\dfrac{1}{9}y^2\)

a) \(=4x^2-12x+9\)

b) \(=4x^2+2x+\dfrac{1}{4}\)

c) \(=4x^2-\dfrac{4}{3}x+\dfrac{1}{9}\)

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

e) \(=\left(3-\dfrac{x}{2}\right)\left(9+\dfrac{3x}{2}+\dfrac{x^2}{4}\right)\)

f) \(=\left(125-4x\right)\left(125^2+500x+16x^2\right)\)

b) ( x - √7y )² = x² - 2x.7y + ( √7y )² 

= x² - 14xy + 7y

a) \(\left(3x-2\right)^2=\left(3x\right)^2-2.3x.2+2^2=9x^2-12x+4\)

b) \(\left(\dfrac{x}{3}+y^3\right)^2=\left(\dfrac{x}{3}\right)^2+2\dfrac{x}{3}y^3+\left(y^3\right)^2=\dfrac{x^2}{9}+\dfrac{2}{3}xy^3+y^6\)

c) \(9x^2-225=\left(3x\right)^2-\left(15\right)^2=\left(3x-15\right)\left(3x+15\right)\)

d) \(\left(2x-3y\right)^3=\left(2x\right)^3-3\left(2x\right)^23y+3.2x\left(3y\right)^2-\left(3y\right)^3=8x^3-3.4x^2.3y+6x.9y^2-27y^3=8x^3-36x^2y+54xy^2-27y^3\)

e) \(\left(2x^2+\dfrac{3}{2}\right)^3=\left(2x^2\right)^3+3\left(2x^2\right)^2\dfrac{3}{2}+3.2x^2\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3=8x^6+3.4x^4.\dfrac{3}{2}+6x^2.\dfrac{9}{4}+\dfrac{27}{8}=8x^6+18x^4+\dfrac{27}{2}x^2+\dfrac{27}{8}\)

f) \(\left(-2xy^2+\dfrac{1}{2}x^3y\right)^3=\left(-2xy^2\right)+3\left(-2xy^2\right)^2\dfrac{1}{2}x^3y+3\left(-2xy^2\right)\left(\dfrac{1}{2}x^3y\right)^2+\left(\dfrac{1}{2}x^3y\right)^3=-8x^3y^6+3.4x^2y^4.\dfrac{1}{2}x^3y-6xy^2.\dfrac{1}{4}x^6y^2+\dfrac{1}{8}x^9y^3=-8x^3y^6+6x^5y^5-\dfrac{3}{2}x^7y^4+\dfrac{1}{8}x^9y^3\)

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

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

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

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

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

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

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

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

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

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

\(a,=x^3+3x^2+3x+1\\ b,=8x^3+36x^2+54x+27\\ c,=x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}\\ d,=x^6-6x^4+12x^2-8\\ e,=8x^3-36x^2y+54xy^2-27y^3\)