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

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

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

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

Giải:

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

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

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

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

Vậy ...

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

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

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

\(=x^2-4xy+4y^2+2x-4y+1\)

Vậy ...

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

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

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

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

Vậy ...

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

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

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) \({\left( {2x + 1} \right)^4} = {\left( {2x} \right)^4} + 4.{\left( {2x} \right)^3}{.1^1} + 6.{\left( {2x} \right)^2}{.1^2} + 4.\left( {2x} \right){.1^3} + {1^4} = 16{x^4} + 32{x^3} + 24{x^2} + 8x + 1\)

b) \(\begin{array}{l}{\left( {3y - 4} \right)^4} = {\left[ {3y + \left( { - 4} \right)} \right]^4} = {\left( {3y} \right)^4} + 4.{\left( {3y} \right)^3}.\left( { - 4} \right) + 6.{\left( {3y} \right)^2}.{\left( { - 4} \right)^2} + 4.{\left( {3y} \right)^1}{\left( { - 4} \right)^3} + {\left( { - 4} \right)^4}\\ = 81{y^4} - 432{y^3} + 864{y^2} - 768y + 256\end{array}\)

c) \({\left( {x + \frac{1}{2}} \right)^4} = {x^4} + 4.{x^3}.{\left( {\frac{1}{2}} \right)^1} + 6.{x^2}.{\left( {\frac{1}{2}} \right)^2} + 4.x.{\left( {\frac{1}{2}} \right)^3} + {\left( {\frac{1}{2}} \right)^4} = {x^4} + 2{x^3} + \frac{3}{2}{x^2} + \frac{1}{2}x + \frac{1}{{16}}\)

d) \(\begin{array}{l}{\left( {x - \frac{1}{3}} \right)^4} = {\left[ {x + \left( { - \frac{1}{3}} \right)} \right]^4} = {x^4} + 4.{x^3}.{\left( { - \frac{1}{3}} \right)^1} + 6.{x^2}.{\left( { - \frac{1}{3}} \right)^2} + 4.x.{\left( { - \frac{1}{3}} \right)^3} + {\left( { - \frac{1}{3}} \right)^4}\\ = {x^4} - \frac{4}{3}{x^3} + \frac{2}{3}{x^2} - \frac{4}{27}x + \frac{1}{{81}}\end{array}\)