a/ đk: x≥ 0; x # 25
A = \(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{5-\sqrt{x}}-\dfrac{3x+4\sqrt{x}-5}{x-4\sqrt{x}-5}\)
= \(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{\sqrt{x}-5}-\dfrac{3x+4\sqrt{x}-5}{x-4\sqrt{x}-5}\)
= \(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-5\right)}+\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+1\right)}-\dfrac{3x+4\sqrt{x}-5}{x-4\sqrt{x}-5}\)
=\(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-5\right)+\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)-3x-4\sqrt{x}+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-5\right)}\)
= \(\dfrac{x-3\sqrt{x}-10+x+4\sqrt{x}+3-3x-4\sqrt{x}+5}{x-4\sqrt{x}-5}\)
= \(\dfrac{-x-3\sqrt{x}-2}{x-4\sqrt{x}-5}\)
b/ A > - 2
<=> \(\dfrac{-x-3\sqrt{x}-2}{x-4\sqrt{x}-5}>-2\)
<=> \(\dfrac{-x-3\sqrt{x}-2}{x-4\sqrt{x}-5}+2>0\)
<=> \(\dfrac{-x-3\sqrt{x}-2+2x-8\sqrt{x}-10}{x-4\sqrt{x}-5}>0\)
<=> \(\dfrac{x-11\sqrt{x}-12}{x-4\sqrt{x}-5}>0\)
<=> \(\left\{{}\begin{matrix}x-11\sqrt{x}-12>0\\x-4\sqrt{x}-5>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>144\\x>25\end{matrix}\right.\Leftrightarrow}x>144}\)(1)
hoặc
\(\left\{{}\begin{matrix}x-11\sqrt{x}-12< 0\\x-4\sqrt{x}-5< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0\le x< 144\\0\le x< 25\end{matrix}\right.\Rightarrow}0\le x< 25}\)(2)
từ (1);(2) => 0 ≤ x < 25 và x > 144
hay x ∈ [0;25) \(\cup\) (144; +\(\infty\))
