1, \(x^2+2xy+y^2=\left(x+y\right)^2\)

2, \(4x^2+12x+9=\left(2x\right)^2+2\cdot3\cdot2x+3^2=\left(2x+3\right)^2\)

3, \(x^2+5x+\dfrac{25}{4}=x^2+2\cdot\dfrac{5}{2}\cdot x+\left(\dfrac{5}{2}\right)^2=\left(x+\dfrac{5}{2}\right)^2\)

4, \(16x^2-8x+1=\left(4x\right)^2-2\cdot4x\cdot1+1^2=\left(4x-1\right)^2\)

5, \(x^2+x+\dfrac{1}{4}=x^2+2\cdot\dfrac{1}{2}\cdot x+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)

1: =(x+y)^2

2: =(2x+3)^2

3: =(x+5/2)^2

4: =(4x-1)^2

5: =(x+1/2)^2

6: =(x-3/2)^2

7: =(x+1)^3

8: =(1/2x+1)^2

9: =(3y-1/3)^3

10: =(2x+y)^3

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

7, \(x^3+3x^2+3x+1=x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=\left(x+1\right)^3\)

8, \(\dfrac{x^2}{4}+x+1=\left(\dfrac{x}{2}\right)^2+2\cdot\dfrac{x}{2}\cdot1+1^2=\left(\dfrac{x}{2}+1\right)^2\)

9, \(27y^3-9y^2+y-\dfrac{1}{27}=\left(3y\right)^3-3\cdot\left(3y\right)^2\cdot\dfrac{1}{3}+3\cdot3y\cdot\left(\dfrac{1}{3}\right)^2-\left(\dfrac{1}{3}\right)^3=\left(3y-\dfrac{1}{3}\right)^3\)

10, \(8x^3+12x^2y+6xy^2+y^3=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2+y^3=\left(2x+y\right)^3\)

a)
=(x-2)³
b)\(\left(2-x\right)^3\)
c)\(\left(x+\dfrac{1}{3}\right)^3\)
d)\(\left(\dfrac{x}{2}+y\right)^3\)
e)
\(=\left(x-1\right)^2\left(x-1-15\right)+25\left[3\left(x-1\right)-5\right]\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-3-5\right)\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-8\right)\)
 

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

a. $x^2+4x+4$

$=x^2+2\cdot x\cdot2+2^2$

$=(x+2)^2$

b. $x^2-6xy+9y^2$

$=x^2-2\cdot x\cdot3y+(3y)^2$

$=(x-3y)^2$

c. $4x^2+12x+9$

$=(2x)^2+2\cdot2x\cdot3+3^2$

$=(2x+3)^2$

d. $x^2-x+\dfrac14$

$=x^2-2\cdot x\cdot \dfrac12+\Bigg(\dfrac12\Bigg)^2$

$=\Bigg(x-\dfrac12\Bigg)^2$

`x^2 +4x+4`

`=x^2+2*x*2+2^2`

`=(x+2)^2`

__

`x^2-6xy+9y^2`

`=x^2 - 2*x*3y+(3y)^2`

`=(x-3y)^2`

__

`4x^2 +12x+9`

`=(2x)^2 +2*2x*3+3^2`

`=(2x+3)^2`

__

`x^2-x+1/4`

`=x^2 - 2*x*1/2 +(1/2)^2`

`=(x+1/2)^2`

xl chuyển hộ mk "bình phương" thành "lập phương" nha

\(=\left(\dfrac{2}{3}a\right)^3-3.\left(\dfrac{2}{3}\right)^2a^2.2b+3.\dfrac{2}{3}a.4b^2-\left(2b\right)^3=\left(\dfrac{2}{3}a-2b\right)^3\)

\(a,=\left(x+\dfrac{5}{2}\right)^2\\ b,=\left(2x+3y\right)^2\\ c,=a^2+b^2+c^2+2ab-2bc-2ac\\ d,=\left(4x-1\right)^2\\ e,=a^2+b^2+c^2+2ab+2bc+2ac\\ f,=a^2+b^2+c^2-2ab+2bc-2ac\)

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

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

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

a) \(x^2+2x+1\)

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

b) \(9-24x+16x^2\)

\(=\left(3-4x\right)^2\)

c) \(4x^2+\dfrac{1}{4}+2x\)

\(=4x^2+2x+\dfrac{1}{4}\)

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

2:

a: a^2-b^2

b: (a-b)^2

c: 1/2(a^2+b^2)

d: 1/2(a^3+b^3)