mk ghi đáp án, ko phân tích đc thì IB mk

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

b) \(4a^4-4a^2b^2+b^4=\left(2a^2-b^2\right)^2\)

c)  \(x^6+y^2-2x^3y=\left(x^3-y\right)^2\)

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

e)  \(25x^4-10x^2y^2+y^4=\left(5x^2-y^2\right)^2\)

f) \(-a^2-2a-1=-\left(a+1\right)^2\)

g)  \(27b^3-8a^3=\left(3b-2a\right)\left(9b^2+6ab+4a^2\right)\)

h)  \(x^3+9x^2y+27xy^2+27y^3=\left(x+3y\right)^3\)

i) \(16x^2-9\left(x+y\right)^2=\left(x-3y\right)\left(7x+3y\right)\)

\(a,=x^2+x+\dfrac{1}{4}\\ b,=4x^2+2x+\dfrac{1}{4}\\ c,=x^2-2+\dfrac{1}{x^2}\\ d,=4x^2+\dfrac{8}{3}x+\dfrac{4}{9}x^2\\ e,=a^2-1\\ f,=25x^4-4\)

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

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

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

\(d,\left(\dfrac{2x+2}{3x}\right)^2=\dfrac{\left(2x+2\right)^2}{9x^2}=\dfrac{4x^2+8x+4}{9x^2}\)

\(e,\left(a-1\right).\left(a+1\right)=a^2-1\)

\(f,\left(5x^2-2\right).\left(5x^2+2\right)=25x^4-4\)

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

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

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

d. \(\left(2x+\dfrac{2}{3x}\right)^2=4x^2+\dfrac{8}{3}+\dfrac{4}{9x^2}\)

e. (a - 1)(a + 1) = a² - 1

f. (5x² - 2)(5x² + 2) = 25x⁴ - 4 

g. (2a - 3)(2a + 3) = 4a² - 9

a) ta có : \(x^2+6xy+9y^2=x^2+2.x.3y+\left(3y\right)^2=\left(x+3y\right)^2\)

b) ta có : \(4a^4-4a^2b^2+b^4=\left(2a^2\right)^2-2.2a^2.b^2+\left(b^2\right)^2=\left(2a^2-b^2\right)^2\)

c) ta có : \(x^6+y^2-2x^3y=\left(x^3\right)^2-2.x^3.y+y^2=\left(x^3-y^2\right)^2\)

d) ta có : \(\left(x+y\right)^3-\left(x-y\right)^3=\left(x+y-x+y\right)\left(\left(x+y\right)^2+2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right)\)

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

\(=2y\left(4x^2\right)=8x^2y\)

e) ta có : \(25x^4-10x^2y^2+y^4=\left(5x^2\right)^2-2.5x^2.y^2+\left(y^2\right)^2=\left(5x^2-y^2\right)^2\)

f) ta có : \(-a^2-2a-1=-\left(a^2+2a+1\right)=-\left(a+1\right)^2\)

g) ta có : \(27b^3-8a^3=\left(3b\right)^3-\left(2a\right)^3=\left(3b-2a\right)\left(9b^2+6ab+4a^2\right)\)

i) ta có : \(16x^2-9\left(x+y\right)^2=\left(4x\right)^2-\left(3\left(x+y\right)\right)^2\)

\(=\left(4x-3x-3y\right)\left(4x+3x+3y\right)=\left(x-3y\right)\left(7x+3y\right)\)

Đề này là Toán chứ nhỉ!?

a) (x + 1)³ = x³ + 3x² + 3x + 1

b) (2x + 3)³ = 8x³ + 36x² + 54x + 27

c) \(\left(x+\frac{1}{2}\right)^3=\) \(x^3+\frac{3x^2}{2}+\frac{3x}{4}+\frac{1}{8}\)

d) (x² + 2)³ = x⁶ + 6x⁴ + 12x² + 8

e) (2x + 3y)³ = 8x³ + 36x²y + 54xy² + 27y³

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

Bài 3:

a) \(4x^2+4x+1=\left(2x+1\right)^2\)

b) \(9x^2-12x+4=\left(3x-2\right)^2\)

c) \(ab^2+\dfrac{1}{4}a^2b^4+1=\left(\dfrac{1}{2}ab^2+1\right)^2\)

Phần d bài 3 mk chưa lm dc

Bài 3: 

a: \(4x^2+4x+1=\left(2x+1\right)^2\)

b: \(9x^2-12x+4=\left(3x-2\right)^2\)

c: \(\dfrac{1}{4}a^2b^4+ab^2+1=\left(\dfrac{1}{2}ab^2+1\right)^2\)

d: 

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

2, đề sai 

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

4, \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)

5, \(1000-y^3=\left(10-y\right)=\left(100+10y+y^2\right)\)

tương tự ... 

8, \(8x^3+27y^3=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

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

b: \(4a^4-4a^2b^2+b^4=\left(2a^2-b^2\right)^2\)

\(x^6-2x^3y+y^2=\left(x^3-y\right)^2\)

b: \(\left(x+y\right)^3-\left(x-y\right)^3\)

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

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

\(25x^4-10x^2y^2+y^4=\left(5x^2-y^2\right)^2\)

\(-a^2-2a-1=-\left(a+1\right)^2\)