1a) ĐKXĐ:

$\begin{cases}x \geq 0\\x - 1 \geq 0\\1 - \sqrt x \ne 0\end{cases}$

$\Leftrightarrow x > 1$

b) $A = \dfrac{1}{\sqrt x + \sqrt{x -1}} - \dfrac{1}{\sqrt x - \sqrt{x -1}} - \dfrac{\sqrt{x^3} - x}{1 - \sqrt x}$

$= \dfrac{\sqrt x - \sqrt{x -1} - (\sqrt x + \sqrt{x -1})}{(\sqrt x + \sqrt{x -1})(\sqrt x - \sqrt{x -1})} + \dfrac{x(\sqrt x -1)}{\sqrt x -1}$

$= \dfrac{-2\sqrt{x -1}}{x - (x -1)} + x$

$= -2\sqrt{x -1} + x$

$A > 0 \Leftrightarrow -2\sqrt{x -1} + x > 0$

$\Leftrightarrow x > 2\sqrt{x -1}$

$\Leftrightarrow x^2 - 4x + 4 > 0$

$\Leftrightarrow (x - 2)^2 > 0$

Ta có: $(x - 2)^2 \geq 0, \forall x$

$\Rightarrow (x -2)^2 > 0 \Leftrightarrow x \ne 2$

Kết hợp điều kiện xác định, ta được: $x > 1; \, x \ne 2$

2) $(x + \sqrt{x^2 + 4})(y + \sqrt{y^2 + 4}) = 4$

Ta được:

$(x - \sqrt{x^2 + 4})(x + \sqrt{x^2 + 4})(y + \sqrt{y^2 + 4}) = 4(x - \sqrt{x^2 + 4})$

$\Leftrightarrow (x^2 - (x^2 +4))(y + \sqrt{y^2+4}) = 4(x - \sqrt{x^2 + 4})$

$\Leftrightarrow -4(y + \sqrt{y^2+4}) = 4(x - \sqrt{x^2 + 4})$

$\Leftrightarrow y + \sqrt{y^2 + 4} =\sqrt{x^2 + 4} - x$ $(1)$

Tương tự, ta được:

$x + \sqrt{x^2 + 4} = \sqrt{y^2 + 4} - y$ $(2)$

Cộng hai vế $(1)$ và $(2)$ ta được:

$x + y = - x - y$

$\Leftrightarrow 2(x + y) = 0$

$\Leftrightarrow x + y = 0$

Ta có:

$A = x^3 + y^3$

$= (x + y)(x^2 - xy + y^2)$

Do $(x + y) = 0$

nên $(x + y)(x^2 - xy + y^2) = 0$

Hay $A = 0$

3a) $A = \dfrac{1}{\sqrt2 + \sqrt1} + \dfrac{1}{\sqrt3 + \sqrt2} +\cdots+ \dfrac{1}{\sqrt{100} + \sqrt{99}}$

$= \dfrac{\sqrt{2} - \sqrt{1}}{(\sqrt2 + \sqrt1)(\sqrt2 - 1)} + \dfrac{\sqrt3 - \sqrt2}{(\sqrt3 + \sqrt2)(\sqrt3 - \sqrt2)} +\cdots+ \dfrac{\sqrt{100} -\sqrt{99}}{(\sqrt{100} + \sqrt{99})(\sqrt{100} - \sqrt{99})}$

$= \sqrt2 - \sqrt1 + \sqrt3 - \sqrt2 +\cdots+ \sqrt{100} - \sqrt{99}$

$= - \sqrt1 + \sqrt{100}$

$=10 - 1 = 9$

b) $B = \dfrac{1}{2\sqrt1 + 1\sqrt2} + \dfrac{1}{3\sqrt2 + 2\sqrt3} + \cdots + \dfrac{1}{100\sqrt{99} + 99\sqrt{100}}$

$= \dfrac{1}{\sqrt2.\sqrt1(\sqrt2 + \sqrt1)} + \dfrac{1}{\sqrt3.\sqrt2(\sqrt3 + \sqrt2)} +\cdots+ \dfrac{1}{\sqrt{100}.\sqrt{99}(\sqrt{100} + \sqrt{99})}$

$= \dfrac{\sqrt2 - \sqrt1}{\sqrt2.\sqrt1} + \dfrac{\sqrt3 - \sqrt2}{\sqrt3.\sqrt2} + \cdots+ \dfrac{\sqrt{100} - \sqrt{99}}{\sqrt{100}.\sqrt{99}}$

$= \dfrac{1}{\sqrt1} - \dfrac{1}{\sqrt2} + \dfrac{1}{\sqrt2} - \dfrac{1}{\sqrt3} +\cdots + \dfrac{1}{\sqrt{99}} - \dfrac{1}{\sqrt{100}}$

$= 1 - \dfrac{1}{10} = \dfrac{9}{10}$