Đặt \(x^2=a,y^2=b\) ta có :
\(3a^2+5ab+2b^2+2b\)
\(=3a^2+3ab+2ab+2b^2+2b\)
\(=3a\left(a+b\right)+2b\left(a+b\right)+2b\)
\(=\left(a+b\right)\left(3a+2b\right)+2b\)
Mà theo đề bài \(x^2+y^2=2\) \(\Leftrightarrow a+b=2\).
\(\Leftrightarrow2\left(3a+2b\right)+2b\)
\(=2\left(2a+2b\right)+2a+2b=8+4=12\)