a. ĐKXĐ: $x \geq 0$; $x \neq 1$
$P = \dfrac{\sqrt{x}}{\sqrt{x} - 1} + \dfrac{3}{\sqrt{x} + 1} - \dfrac{6\sqrt{x} - 4}{x - 1} =$
$= \dfrac{(\sqrt{x}(\sqrt{x} + 1) + 3(\sqrt{x} - 1) - 6\sqrt{x} - 4}{(\sqrt{x} - 1)(\sqrt{x} + 1)} =$
$= \dfrac{x + \sqrt{x} + 3\sqrt{x} - 3 - 6\sqrt{x} + 4}{(\sqrt{x} - 1)(\sqrt{x} + 1)} = \dfrac{x - 2\sqrt{x} + 1}{(\sqrt{x} - 1)(\sqrt{x} + 1)} = \dfrac{(\sqrt{x} - 1)^2}{(\sqrt{x} - 1)(\sqrt{x} + 1)} = \dfrac{\sqrt{x} - 1}{\sqrt{x} + 1}$
b. $P = \dfrac{3}{4} \to \dfrac{\sqrt{x} - 1}{\sqrt{x} + 1} = \dfrac{3}{4}$
Quy đồng - Khử mẫu ta được:
$3\sqrt{x} + 3 = 4\sqrt{x} - 4$
$<=> \sqrt{x} = 7 <=> x = 49$ (Thoã mãn)
Vậy với $x = 49$ thì $P = \dfrac{3}{4}$
c. $P < \dfrac{1}{2} \to \dfrac{\sqrt{x} - 1}{\sqrt{x} + 1} < \dfrac{1}{2}$
$<=> ...$
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