$B\,=(x+1)^2+(2x-3)^2\\\quad =x^2+2x+1+4x^2-12x+9\\\quad =5x^2-10x+10\\\quad =5(x^2-2x+2)\\\quad =5(x^2-2x+1+1)\\\quad =5(x-1)^2+5$
Nhận thấy: $(x-1)^2\ge 0$
$↔5(x-1)^2\ge 0\\↔B\ge 5\\→\min B=5$
$→$ Dấu "=" xảy ra khi $x-1=0$
$↔x=1$
Vậy $\min B=5$ khi $x=1$