Question

Viết dưới dạng tích:

a) x² +6x +9

b) 4x² -4x +1

c) 1/8 -x³

d) 64x³ -1/27

Answer 1

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

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

c, \(\dfrac{1}{8}-x^3=\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{64}+\dfrac{1}{2}x+x^2\right)\)

b, \(64x^3-\dfrac{1}{27}=\left(4x-\dfrac{1}{3}\right)\left(16x^2+\dfrac{4}{3}x+\dfrac{1}{9}\right)\)

Answer 2

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

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

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

b, \(64x^3-\dfrac{1}{27}=\left(4x-\dfrac{1}{3}\right)\left(16x^2+\dfrac{4}{3}x+\dfrac{1}{9}\right)\)

Answer 3

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

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

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

d, \(64x^3-\dfrac{1}{27}=\left(4x-\dfrac{1}{3}\right)\left(16x^2+\dfrac{4}{3}x+\dfrac{1}{9}\right)\)