Question

[Đỗ Khánh Linh]

phân tích đa thức  A=(x^2+x)^2+2(x^2+x)+1 ; B= (x-a)^4-(x+a)^4 thành phân tử

Answer 1

[Nguyễn Huy Tú]

\(A=\left(x^2+x\right)^2+2\left(x^2+x\right)+1=\left(x^2+x+1\right)^2\)

\(B=\left(x-a\right)^4-\left(x+a\right)^4=\left[\left(x-a\right)^2\right]^2-\left[\left(x+a\right)^2\right]^2\)

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

\(=\left(x-a-x-a\right)\left(x-a+x+a\right)\left(x^2-2xa+a^2+x^2+2ax+a^2\right)\)

\(=-2a.2x\left(2x^2+2a^2\right)=-8ax\left(x^2+a^2\right)\)

Answer 2

[Capheny Bản Quyền]

A=(x^2+x)^2+2(x^2+x)+1 =(x^2+x+1)^2