\(a,\left(2x+y+3\right)^2=4x^2+y^2+9+4xy+12x+6y\)

\(b,\left(x-2y+1\right)^2=x^2+4y^2+1-4xy+2x-4y\)

\(c,\left(x^2-2xy^2-3\right)^2=x^4+2x^2y^4+9-4x^3y^2-6x^2+12xy^2\)

a,\(\left(x^2+2xy\right)^3=\left(x^2\right)^3+3.\left(x^2\right)^2.2xy+3.\left(2xy\right)^2.x^2+\left(2xy\right)^3\)

\(=x^6+6x^5y+12x^4y^2+8x^3y^3\)

b,\(\left(3x^2-2y\right)^3=\left(3x^2\right)^3-3.\left(3x^2\right)^2.2y+3.\left(2y\right)^2.3x^2-\left(2y\right)^3\)

\(=27x^6-54x^4y+36y^2x^2-8y^3\)

c,\(\left(2x^3-y^2\right)^3=8x^9-12x^6y^2+6x^3y^4-y^6\)

a) \(\left(3x^2-2y^3\right)^2\)

\(=\left(3x^2\right)^2-2\cdot3x^2\cdot2y^3+\left(2y^3\right)^2\)

\(=9x^4-12x^2y^3+4y^6\)

b) \(\left(-2x^2-3\right)^2\)

\(=\left(-2x^2\right)^2-2\cdot\left(-2x^2\right)\cdot3+3^2\)

\(=4x^4+12x^2+9\)

a) \(\left(x^2+2xy\right)^3\)

\(=\left(x^2\right)^3+3\left(x^2\right)^22xy+3x^2\left(2xy\right)^2+\left(2xy\right)^3\)

\(=x^6+6x^5y+12x^4y^2+8x^3y^3\)

b) \(\left(3x^2-2y\right)^3\)

\(=\left(3x^2\right)^3-3\left(3x^2\right)^22y+3.3x^2\left(2y\right)^2-\left(2y\right)^3\)

\(=27x^6-54x^4y+36x^2y^2-8y^3\)

c) \(\left(2x^3-y^2\right)^3\)

\(=\left(2x^3\right)^3-3\left(2x^3\right)^2y^2+3.2x^3\left(y^2\right)^2-\left(y^2\right)^3\)

\(=8x^9-12x^6y^2+6x^3y^4-y^6.\)

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

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

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) \(\left(2x^2-1\right)^2=\left(2x^2\right)^2-2.2x^2.1+1^2\)

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

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

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

Chúc bn hc tốt!

Giải:

a) \(\left(3x^2-2y^3\right)^2\)

\(=\left(3x^2\right)^2-2.3x.2y+\left(2y^3\right)^2\)

\(=9x^4-12xy+4y^6\)

Vậy ...

b) \(\left(-2x^2-3\right)^2\)

\(=\left(-2x^2\right)^2-2.2x^2.3+3^2\)

\(=4x^4-12x^2+9\)

Vậy ...

a. (2x+3y)²= (2x)²+2.2x.3y+(3y)²

=4x²+12xy+9y²

b. 2(\(\dfrac{1}{2}\)x²+y)(x²-2y)

=(x²+2y)(x²-2y)

=x⁴-4y²

c, (x+y+z)²= [(x+y)+z]²

=(x+y)²+2(x+y)z+z²

=x²+2xy+y²+2xz+2yz+z²

=x²+y²+z²+2xy+2yz+2xz

a,\(\left(2x-1\right)\left(4x^2+2x+1\right)=\left(2x-1\right)\left[\left(2x\right)^2+2x.1+1^2\right]\)

\(=\left(2x\right)^3-1=8x^3-1\)

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

\(=x^2+2.x.2y+\left(2y\right)^2-z^2=x^2+4xy+4y^2-z^2\)

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

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

`=(2x)^3-1^3`

`=8x^3-1`

Áp dụng HĐT:`A^3-B^3=(A-B)(A^2+AB+B^2)`

`b)(x+2y+z)(x+2y-z)`

`=[(x+2y)+z][(x+2y)-z]`

`=(x+2y)^2-z^2`

`=x^2+2.x.2y+(2y)^2-z^2`

`=x^2+4xy+4y^2-z^2`

Áp dụng HĐT:`A^2-B^2=(A+B)(A-B)`

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