Đáp án:
$B=-y-\dfrac{3}{4}$
Giải thích các bước giải:
$B=\dfrac{y}{2}+\dfrac{3}{4}\sqrt{1-4y+4y^2}-\dfrac{3}{2}\left(y \le \dfrac{1}{2}\right)\\ =\dfrac{y}{2}+\dfrac{3}{4}\sqrt{(1-2y)^2}-\dfrac{3}{2}\\ =\dfrac{y}{2}+\dfrac{3}{4}|1-2y|-\dfrac{3}{2}\\ =\dfrac{y}{2}+\dfrac{3}{4}(1-2y)-\dfrac{3}{2}\\ =\dfrac{y}{2}+\dfrac{3}{4}-\dfrac{3y}{2}-\dfrac{3}{2}\\ =-y-\dfrac{3}{4}$
