$A=x^2+3x+5\\=x^2+2.x.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{11}{4}\\=\left(x+\dfrac{3}{2}\right)^2+\dfrac{11}{4}\\\left(x+\dfrac{3}{2}\right)^2≥0\\→\left(x+\dfrac{3}{2}\right)^2+\dfrac{11}{4}≥\dfrac{11}{4}\\→A≥\dfrac{11}{4}\\→\min A=\dfrac{11}{4}$
$→$ Dấu "=" xảy ra khi $x+\dfrac{3}{2}=0$
$↔x=-\dfrac{3}{2}$
Vậy $\min A=\dfrac{11}{4}$ khi $x=-\dfrac{3}{2}$
$\\\\$
$B=2x^2+4x-5\\=2\left(x^2+2x-\dfrac{5}{2}\right)\\=2\left(x^2+2x+1-\dfrac{7}{2}\right)\\=2(x+1)^2-7\\(x+1)^2≥0\\→2(x+1)^2≥0\\→2(x+1)^2-7≥-7\\→B≥-7\\→\min B=-7$
$→$ Dấu "=" xảy ra khi $x+1=0$
$↔x=-1$
Vậy $\min B=-7$ khi $x=-1$
