Question

phân tích đa thức thành nhân tử

a.x¹⁶-1

b.x⁶+y⁶

Answer 1

a) \(x^{16}-1=\left(x^8\right)^2-1^2=\left(x^8-1\right)\left(x^8+1\right)=\left(\left(x^4\right)^2-1^2\right)\left(x^8+1\right)\)

\(=\left(x^4-1\right)\left(x^4+1\right)\left(x^8+1\right)=\left(\left(x^2\right)^2-1^2\right)\left(x^4+1\right)\left(x^8+1\right)\)

\(=\left(x^2-1^2\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\)

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

Answer 2

\(a,\\ x^{16}-1=\left(x^8\right)^2-1^2=\left(x^8-1\right)\left(x^8+1\right)\)\(b,\\ x^6+y^6=\left(x^2\right)^3+\left(y^2\right)^3=\left(x^2+y^2\right)\left(\left(x^2\right)^2-\left(xy\right)^2+\left(y^2\right)^2\right)=\left(x^2+y^2\right)\left(x^4-\left(xy\right)^2+y^4\right)\)