\(a,\left(x+2\right)^2=x^2+4x+4\\ b,\left(x-1\right)^2=x^2-2x+1\\ c,\left(x^2+y^2\right)^2=x^4+2x^2y^2+y^4\)

a) = x² + 4x + 4

b) = x² - 2x + 1

c) x⁴ + 2x²y² + y⁴

( A   –   B ) 3   =   ( A   +   ( - B ) ) 3       =   A 3   +   3 . A 2 . ( - B )   +   3 . A . ( - B ) 2   +   ( - B ) 3       =   A 3   –   3 A 2 B   +   3 A B 2   –   B 3     = >   ( A   –   B ) 3   =   A 3   –   3 A 2 B   –   3 A B 2   +   B 3

là sai

Đáp án cần chọn là: B

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

\(8)\)  \(x^2-\left(2y\right)^2=\left(x-2y\right)\left(x+2y\right)\)

\(9)\)  \(\left(2x\right)^2-y^2=\left(2x-y\right)\left(2x+y\right)\)

a)
x³+3x²+3x+1
b)8x³+18x²+54x+27
c)x³+3636x²+3636x+1780360128
d)x⁶-6x²+12x-8
e)
8x³-36x²y+54xy²-27y³

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

\(27x^3-1\) =\(\left(3x\right)^3-1^3\)

Nhớ tick và theo dõi mình nha!

\(a,=x^3+3x^2+3x+1\\ b,=8x^3+36x^2+54x+27\\ c,=x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}\\ d,=x^6-6x^4+12x^2-8\\ e,=8x^3-36x^2y+54xy^2-27y^3\)

(1𝑦/3+3)^3

(𝑦/3+3)^3

(𝑦/3+3⋅3/3)^3

(𝑦+3⋅3/3)^3

(𝑦+9/3)^3

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

a) \(=4x^2-12x+9\)

b) \(=4x^2+2x+\dfrac{1}{4}\)

c) \(=4x^2-\dfrac{4}{3}x+\dfrac{1}{9}\)

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

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

f) \(=\left(125-4x\right)\left(125^2+500x+16x^2\right)\)