a) \(25x^2-20xy+4y^2\)

\(=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2\)

\(=\left(5x-2y\right)^2\)

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

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

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

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

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

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

d) \(-6xy+x^2+9y^2\)

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

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

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

Mik k gửi đc hình

a) \(x^2-6x+9=x^2-2.3.x+3^2=\left(x-3\right)^2\)

b)\(x^2+4x+4=x^2+2.2.x+2^2=\left(x+2\right)^2\)

c)\(4x^2+4x+1=\left(2x\right)^2+2.2x.1+1^2=\left(2x+1\right)^2\)

d)\(4x^2+12xy+9y^2=\left(2x\right)^2+2.2x.3y+\left(3y\right)^2=\left(2x+3y\right)^2\)

e)\(x^2-8x+16=x^2-2.4.x+4^2=\left(x-4\right)^2\)

a) x² -6x +9 = (x-3)²                                       

b) x²+4x +4= (x+2)²

c) 4x²+4x+1= (2x+1)²

d) 4x²+12xy+9y² = (2x+3y)²

e) x²-8x+16 = (x-4)²

Đây chính là hằng đẳng thức nhé bn....

a) \(x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2\)

b) \(x^2+12xy+36y^2=\left(x+6y\right)^2\)

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

d) Không phải hằng đẳng thức \(\left(x^2-2x+4=\left(x-1\right)^2+3\right)\)

e) \(25x^2+4y^2-20xy=\left(5x-2y\right)^2\)

a) ( 2 x   +   1 ) 2 .                     b) ( 3 x   –   2 ) 2 .

c) 1 2 ab 2 + 1 2 .                  d) ( 4 uv ²   –   1 ) 2 .

a) \(x^2+4x+4\)

\(=x^2+2\cdot2\cdot x+2^2\)

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

b) \(4x^2-4x+1\)

\(=\left(2x\right)^2-2\cdot2x\cdot1+1^2\)

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

c) \(x^2-x+\dfrac{1}{4}\)

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

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

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

\(=\left[2\left(x+y\right)\right]^2-2\cdot2\left(x+y\right)\cdot1+1^2\)

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

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

a) \(A=8x^3+12x^2y+6xy^2+y^3=\left(2x+y\right)^3\)

b) \(B=x^3+3x^2+3x+1=\left(x+1\right)^3\)

c) \(C=x^3-3x^2+3x-1=\left(x-1\right)^3\)

d)  \(D=27+27y^2+9y^4+y^6=\left(3+y^2\right)^3\)

\(1,\\ a,=x^2+2xy+y^2\\ b,=x^2-4xy+4y^2\\ c,=x^2y^4-1\\ d,=\left[\left(x-y\right)\left(x+y\right)\right]^2=\left(x^2-y^2\right)^2=x^4-2x^2y^2+y^4\\ 2,\\ a,=\left(x+2\right)^2\\ b,=\left(3x-2\right)^2\\ c,=\left(\dfrac{x}{2}+1\right)^2\\ d,=\left(x+y-2\right)^2\)

Bài 1 em dùng HĐT nha

Bài 2:

a. x² + 4x + 4

= x² + 2.2.x + 2²

= (x + 2)²

b. 9x² - 12x + 4

= (3x)² - 3x.2.2 + 2²

= (3x - 2)²

c. \(\dfrac{x^2}{4}+x+1\)

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

= \(\left(\dfrac{x}{2}+1\right)^2\) 

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

a/ 9x²-12xy+4y² = (3x - 2y)²

b/ 25x²-10x+1 = (5x - 1)²

c/ 9x²-12x+4 = (3x - 2)²

d/ 4x²+20x+25 = (2x + 5)²

e/ x⁴-4x²+4 = (x² - 2)²

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

b/\(\left(5x-1\right)^2\)

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

d/\(\left(2x+5\right)^2\)

e/\(\left(x-2\right)^2\)