Question

Phân tích đa thức thành nhân tử:

a)1/64x^6-125y^3

b)x^6+1

c)x^6-y^6

d)x^9+1

e)x^12-y^4

(Các bn giúp mik vs)

Answer 1

a)\(\frac{1}{64}x^6-125y^3=\left(\frac{1}{4}x^2\right)^3-\left(5y\right)^3\)\(=\left(\frac{1}{4}x^2-5y\right)\left(\frac{1}{16}x^4+\frac{5}{4}x^2y+25y^2\right)\)

b)\(x^6+1=\left(x^2\right)^3+1^3=\left(x^2+1\right)\left(x^4+x^2+1\right)\)

c)\(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3+y^3\right)\left(x^3-y^3\right)\)

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

d)\(x^9+1=\left(x^3\right)^3+1=\left(x^3+1\right)\left(x^6-x^3+1\right)\)

\(=\left(x+1\right)\left(x^2-x+1\right)\left(x^6-x^3+1\right)\)

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