Đặt \(x^2=a\); \(y^2=b\)
Theo đề bài, ta có: a+b=1
Ta có: \(3x^4+5x^2y^2+2y^4+2y^2\)
\(=3a^2+5ab+2b^2+2b\)
\(=\left(3a^2+3ab\right)+\left(2ab+2b^2\right)+2b\)
\(=3a\left(a+b\right)+2b\left(a+b\right)+2b\)
\(=\left(a+b\right)\left(3a+2b\right)+2b\)
\(=\left(3a+2b\right)\cdot1+2b\)
\(=3a+2b+2b=3a+4b\)
Đề sai rồi bạn