a, \(\left(3-x\right)^2=9-6x+x^2\)

b, \(\left(x-\frac{1}{2}\right)^2=x^2-x+\frac{1}{4}\)

c, \(\left(2x+y\right)^2=4x^2+4xy+y^2\)

a) Ta có: \(\left(x-3\right)^3\)

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

\(=x^3-9x^2+27x^2-27\)

b) Ta có: \(\left(2x-3\right)^3\)

\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2-3^3\)

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

c) Ta có: \(\left(x-\frac{1}{2}\right)^3\)

\(=x^3-3\cdot x^2\cdot\frac{1}{2}+3\cdot x\cdot\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^3\)

\(=x^3-\frac{3}{2}x^2+\frac{3}{4}x-\frac{1}{8}\)

d) Ta có: \(\left(x^2-2\right)^3\)

\(=\left(x^2\right)^3-3\cdot\left(x^2\right)^2\cdot2+3\cdot x^2\cdot2^2-2^3\)

\(=x^6-6x^4+12x^2-8\)

e) Ta có: \(\left(2x-3y\right)^3\)

\(=\left(2x\right)^3-2\cdot\left(2x\right)^2\cdot3y+2\cdot2x\cdot\left(3y\right)^2-\left(3y\right)^3\)

\(=8x^3-24x^2y+36xy^2-27y^3\)

f) Ta có: \(\left(\frac{1}{2}x-y^2\right)^3\)

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

\(=\frac{1}{8}x^3-\frac{3}{4}x^2y^2+\frac{3}{2}xy^4-y^6\)

a) Ta có: \(\left(x+1\right)^3\)

\(=x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3\)

\(=x^3+3x^2+3x+1\)

b) Ta có: \(\left(2x+3\right)^3\)

\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2+3^3\)

\(=8x^3+3\cdot4x^2\cdot3+27\cdot2x+27\)

\(=8x^3+36x^2+54x+27\)

c) Ta có: \(\left(x+\frac{1}{2}\right)^3\)

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

\(=x^3+x^2+\frac{1}{2}x+\frac{1}{8}\)

d) Ta có: \(\left(x^2+2\right)^3\)

\(=\left(x^2\right)^3+3\cdot\left(x^2\right)^2\cdot2+3\cdot x^2\cdot2^2+2^3\)

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

e) Ta có: \(\left(2x+3y\right)^3\)

\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3y+3\cdot2x\cdot\left(3y\right)^2+\left(3y\right)^3\)

\(=8x^3+36x^2y+54xy^2+27y^3\)

f) Ta có: \(\left(\frac{1}{2}x+y^2\right)^3\)

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

\(=\frac{1}{8}x^3+\frac{3}{4}x^2y^2+\frac{3}{2}xy^4+y^6\)

1) \(\left(\frac{1}{4}+k\right)^2=\frac{1}{16}+\frac{1}{2}k+k^2\)

2) \(\left(2x^2y+\frac{1}{2}xy^2\right)^2=4x^4y^2+2x^3y^3+\frac{1}{4}x^2y^4\) (hẳn đề là như thế này)

3) \(\left(x+\frac{1}{2}y\right)^2=x^2+xy+\frac{1}{4}y^2\)

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

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

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

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

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`

a) \(\left(2x-1\right)\left(4x^2+2x+1\right)=8x^3-1\)

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

a) \(\left(2x-1\right)\left(4x^2+2x+1\right)=\left(2x\right)^3-1^3=8x^3-1\)

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