Question

[Dudũbng Luu]

Phân tích đa thức thành nhân tử bằng phương pháp dùng hằng đẳng thức:

8x^3 - 27y^6

a^3b^3c^3 - 1

64x^3 + 1/8y^3

125 + y^3

a^6 - b^6

4x^2 - 9(3x + 5)^2

Answer 1

[Yukru]

a) \(8x^3-27y^6\)

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

\(=\left(2x-3y^2\right)\left[\left(2x\right)^2+2x.3y+\left(3y\right)^2\right]\)

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

b) \(a^3b^3c^3-1\)

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

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

c) \(64x^3+\dfrac{1}{8}y^3\)

\(=\left(4x\right)^3+\left(\dfrac{1}{2}y\right)^3\)

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

\(=\left(4x+\dfrac{1}{2}y\right)\left(4x^2+2xy+\dfrac{1}{4}y^2\right)\)

d) \(125+y^3\)

\(=5^3+y^3\)

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

e) \(a^6-b^6\)

\(=\left(a^3\right)^2-\left(b^3\right)^2\)

\(=\left(a^3-b^3\right)\left(a^3+b^3\right)\)

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

f) \(4x^2-9\left(3x+5\right)^2\)

\(=\left(2x\right)^2-\left[3\left(3x+5\right)\right]^2\)

\(=\left[2x-3\left(3x+5\right)\right]\left[2x+3\left(3x+5\right)\right]\)

\(=\left(2x-9x-15\right)\left(2x+9x+15\right)\)

\(=\left(-7x-15\right)\left(11x+15\right)\)