Mình chỉ làm được a và b thôi
a)P = ($\frac{2x+3}{x-4}$ -$\frac{\sqrt{x}}{\sqrt{x}-2}$ +$\frac{2}{\sqrt{x}+2}$):$\frac{\sqrt{x}-1}{\sqrt{x}-2}$
P = ($\frac{2x+3}{(\sqrt{x}-2)(\sqrt{x}+2)}$ -$\frac{\sqrt{x}}{\sqrt{x}-2}$ +$\frac{2 {\sqrt{x}+2}$):$\frac{\sqrt{x}-1}{\sqrt{x}-2}$
P = $\frac{2x+3-\sqrt{x}(\sqrt{x}+2) + 2(\sqrt{x}-2)}{(\sqrt{x}+2)(\sqrt{x}-2)}$ : $\frac{\sqrt{x}-1}{\sqrt{x}-2}$
P = $\frac{2x+3-x-2\sqrt{x}+2\sqrt{x}-4}{(\sqrt{x}+2)(\sqrt{x}-2)}$:$\frac{\sqrt{x}-1}{\sqrt{x}-2}$
P = $\frac{x-1}{(\sqrt{x}+2)(\sqrt{x}-2)}$.$\frac{\sqrt{x}-2}{\sqrt{x}-1}$
P = $\frac{1}{\sqrt{x}+2}$
Vậy P = $\frac{1}{\sqrt{x}+2}$ với x ≥ 0, x $\neq$ 1, x $\neq$ 4
b) Để P >$\frac{3}{4}$ thì
$\frac{1}{\sqrt{x}+2}$ > $\frac{3}{4}$
<=>$\frac{1}{\sqrt{x}+2}$>$\frac{3}{4}$
<=>$\frac{4}{4(\sqrt{x}+2)}$>$\frac{3(\sqrt{x}+2)}{4(\sqrt{x}+2)}$
<=>4 > 3($\sqrt{x}$+2)
<=>4 > 3$\sqrt{x}$ + 6
<=>4 - 6 > 3$\sqrt{x}$
<=>2 > 3$\sqrt{x}$
<=> - 3$\sqrt{x}$ > 2
<=> $\sqrt{x}$ < $\frac{-2}{3}$
<=> x < $\frac{4}{9}$
Vậy để P > $\frac{3}{4}$ thì x < $\frac{4}{9}$
