Đề là PTĐTTNT hả

\(6x^2+4x+2=2\left(3x^2+2x+1\right)\)

\(=\left(2\sqrt{2}-3\right)^2\)

\(\left(3a+4\right)^2+\left(4a-1\right)^2+\left(2+5a\right)\left(2-5a\right)=9a^2+24a+16+16a^2-8a+1+4-25a^2=16a+21\)

\(\Rightarrow\dfrac{4}{28}< \dfrac{7x}{28}< \dfrac{16}{28}\)

=> 4 < 7x < 16

Vì x ∈ N

=> x ∈ {1; 2}