Question

3.phân tích về dạng tích

a. x³+64

b.\(\dfrac{1}{27}\)x³ +8y³

c. x³-27y³

d.x⁶-\(\dfrac{1}{27}\)

giải giúp mình bàu này với ạ!

Answer 1

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

b) \(\dfrac{1}{27}x^3+8y^3=\left(\dfrac{1}{3}x\right)^3+\left(2y\right)^3=\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)\)

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

d) \(x^6-\dfrac{1}{27}=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3=\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)\)