Đáp án:
$\min B = -\dfrac14 \Leftrightarrow x =\dfrac14$
Giải thích các bước giải:
$\quad A = \dfrac{\sqrt x}{\sqrt x +1}\qquad (x \geqslant 0)$
$\to B = A(x-1)=\dfrac{\sqrt x(x-1)}{\sqrt x+1}$
$\to B =\dfrac{\sqrt x\left(\sqrt x -1\right)\left(\sqrt x +1\right)}{\sqrt x +1}$
$\to B = \sqrt x\left(\sqrt x -1\right)$
$\to B = x - \sqrt x$
$\to B = \left(x - 2\cdot \dfrac12\cdot \sqrt x +\dfrac14\right) -\dfrac14$
$\to B =\left(\sqrt x -\dfrac12\right)^2 -\dfrac14$
Ta có:
$\quad \left(\sqrt x -\dfrac12\right)^2 \geqslant 0 \quad \forall x\geqslant 0$
$\to \left(\sqrt x -\dfrac12\right)^2 -\dfrac14 \geqslant -\dfrac14$
$\to B \geqslant -\dfrac14$
Dấu $=$ xảy ra $\Leftrightarrow \sqrt x -\dfrac12 =0 \Leftrightarrow x =\dfrac14$
Vậy $\min B = -\dfrac14 \Leftrightarrow x =\dfrac14$
