Đáp án:
$B =\dfrac{\sqrt2}{2}$
Giải thích các bước giải:
`B={\sqrt{x+\sqrt{x^2-y^2}}-\sqrt{x-\sqrt{x^2-y^2}}}/{2\sqrt{x-y}}`
$\to B\sqrt2 = \dfrac{\sqrt{2x + 2\sqrt{x^2 - y^2}} - \sqrt{2x - 2\sqrt{x^2 - y^2}}}{2\sqrt{x-y}}$
$\to B\sqrt2 = \dfrac{\sqrt{x+ y + 2\sqrt{(x+y)(x-y)} + x - y} - \sqrt{x+y - 2\sqrt{(x+y)(x-y)} + x - y}}{2\sqrt{x-y}}$
$\to B\sqrt2 =\dfrac{\sqrt{\left(\sqrt{x+y} + \sqrt{x-y}\right)^2} - \sqrt{\left(\sqrt{x+y} - \sqrt{x-y}\right)^2}}{2\sqrt{x-y}}$
$\to B\sqrt2 = \dfrac{\sqrt{x+y} + \sqrt{x-y} - \left(\sqrt{x+y} - \sqrt{x-y}\right)}{2\sqrt{x-y}}$
$\to B\sqrt2 = \dfrac{2\sqrt{x-y}}{2\sqrt{x -y}}$
$\to B\sqrt2= 1$
$\to B =\dfrac{1}{\sqrt2}=\dfrac{\sqrt2}{2}$
