Question

[Anh Vy]

cú tuii ví dụ 6,7,8 và 9 với, cái vdu 6 là câu a số cuối là số + 9 í, câu b số cuối là + 9x²   và câu d số cuối là + 9y²  

Answer 1

[Nguyễn Lê Phước Thịnh]

9:

a: -x^3+3x^2-3x+1

=(-x)^3+3*(-x)^2*1+3*(-x)*1^2+1^3

=(-x+1)^3

b: z^3-z^2+1/3z-1/27

=z^3-3*z^2*1/3+3*z*(1/3)^2-(1/3)^3

=(z-1/3)^3

c: x^6-3x^4y+3x^2y^2-y^3

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

=(x^2-y)^3

d: =(x-y)^3+3*(x-y)^2*1/3+3*(x-y)*(1/3)^2+(1/3)^3

=(x-y+1/3)^3

Answer 2

[HT.Phong (9A5)]

Ví dụ  9:

a) \(-x^3+3x^2-3x+1\)

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

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

b) \(x^3-x^2+\dfrac{1}{3}x-\dfrac{1}{27}\)

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

\(=\left(x-\dfrac{1}{3}\right)^3\)

c) \(x^6-3x^4y+3x^2y^2-y^3\)

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

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

d) \(\left(x-y\right)^3+\left(x-y\right)^2+\dfrac{1}{3}\left(x-y\right)+\dfrac{1}{27}\)

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

\(=\left(x-y+\dfrac{1}{3}\right)^3\)