Question

[khai]

ai rảnh ko

Answer 1

[Girl]

\(\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\) (*)

Đặt: \(t=x^2+3x-1\) ta có: 

(*)\(=\left(t+2\right)\left(t-2\right)-5=t^2-9=\left(t+3\right)\left(t-3\right)\)

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

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

Answer 2

[khai]

mk vừa làm xong bn ạ