\(B=2x^2+7x+10\\=2(x^2+\dfrac{7}{2}x+5)\\=2(x^2+2.\dfrac{7}{4}x+\dfrac{49}{16}+\dfrac{31}{16})\\=2(x+\dfrac{7}{4})^2+\dfrac{31}{8}\)
Vì \( (x+\dfrac{7}{4})^2\ge 0\\\to 2(x+\dfrac{7}{4})^2\ge 0\\\to B\ge \dfrac{31}{8}\\\to B>0\)
Vậy biểu thức \(B=2x^2+7x+10\) luôn dương