$E=2x^2+3x-1\\=2\left(x^2+\dfrac{3}{2}x-\dfrac{1}{2}\right)\\=2\left(x^2+2.x.\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{17}{16}\right)\\=2\left(x+\dfrac{3}{4}\right)^2-\dfrac{17}{8}$
Nhận thấy $\left(x+\dfrac{3}{4}\right)^2≥0$
$→2\left(x+\dfrac{3}{4}\right)^2≥0\\→E≥-\dfrac{17}{8}\\→\min E=-\dfrac{17}{8}$
$→$ Dấu "=" xảy ra khi $x+\dfrac{3}{4}=0$
$↔x=-\dfrac{3}{4}$
Vậy $\min E=-\dfrac{17}{8}$ khi $x=-\dfrac{3}{4}$
$\\\\$
$F=(2x^2+1)^2+3x^2+10\\=4x^4+4x^2+1+3x^2+10\\=4x^4+7x^2+10\\=(2x^2)^2+2.2x^2.\dfrac{7}{4}+\dfrac{49}{16}+\dfrac{111}{16}\\=\bigg(2x^2+\dfrac{7}{4}\bigg)^2+\dfrac{111}{16}$
Vì $x^2≥0→2x^2≥0$
$→2x^2+\dfrac{7}{4}≥\dfrac{7}{4}\\→\bigg(2x^2+\dfrac{7}{4}\bigg)^2≥\dfrac{49}{16}\\→F≥\dfrac{49}{16}+\dfrac{111}{16}\\→F≥10\\→\min F=10$
$→$ Dấu "=" xảy ra khi $2x^2=0$
$↔x^2=0\\↔x=0$
Vậy $\min F=10$ khi $x=0$
