Question

phân tích thành nhân tử 

a)\(x^6-y^6\)

b)\(x^2+x+\frac{1}{4}\)

Answer 1

a) \(x^6-y^6=\text{(x-y)(y+x)(y^2-xy+x^2)(y^2+xy+x^2)}\)                                                                                                            b)\(x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2\)

Answer 2

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

b)\(x^2+x+\frac{1}{4}=x^2+2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2=\left(x+\frac{1}{2}\right)^2\)

Answer 3

Mình ghi lại nè, câu a mình ghi thiếu mất rồi

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

b)\(x^2+x+\frac{1}{4}=x^2+2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2=\left(x+\frac{1}{2}\right)^2\)