Giải thích các bước giải:

a.ĐKXĐ: $x\ge 0, x\ne 4, x\ne 9$

Ta có:

$F=\dfrac1{2-\sqrt{x}}+\dfrac{\sqrt{x}+3}{\sqrt{x}-3}-\dfrac6{x-5\sqrt{x}+6}$

$\to F=\dfrac{-1}{\sqrt{x}-2}+\dfrac{\sqrt{x}+3}{\sqrt{x}-3}-\dfrac6{(\sqrt{x}-2)(\sqrt{x}-3)}$

$\to F=\dfrac{-(\sqrt{x}-3)+(\sqrt{x}+3)(\sqrt{x}-2)-6}{(\sqrt{x}-2)(\sqrt{x}-3)}$

$\to F=\dfrac{x-9}{(\sqrt{x}-2)(\sqrt{x}-3)}$

$\to F=\dfrac{(\sqrt{x}-3)(\sqrt{x}+3)}{(\sqrt{x}-2)(\sqrt{x}-3)}$

$\to F=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}$

b.Giải $F=1\to \dfrac{\sqrt{x}+3}{\sqrt{x}-2}=1\to \sqrt{x}+3=\sqrt{x}-2\to 3=-2$ vô lý $\to$Vô nghiệm

   Giải $F=-2\to \dfrac{\sqrt{x}+3}{\sqrt{x}-2}=-2$

$\to \sqrt{x}+3=-2\sqrt{x}+4$

$\to 3\sqrt{x}=1$

$\to \sqrt{x}=\dfrac13$

$\to x=\dfrac19$

c.Để $|F|=F$

$\to F\ge 0$

$\to \dfrac{\sqrt{x}+3}{\sqrt{x}-2}\ge 0$

$\to \sqrt{x}-2>0$

$\to \sqrt{x}>2$

$\to x>4$

$\to x>4, x\ne 9$

d.Để $F\in N$

$\to \dfrac{\sqrt{x}+3}{\sqrt{x}-2}\in N$

$\to \dfrac{\sqrt{x}-2+5}{\sqrt{x}-2}\in N$

$\to 1+\dfrac5{\sqrt{x}-2}\in N$

$\to \sqrt{x}-2\in U(5)$

$\to \sqrt{x}-2\in\{1, 5, -1\}$ vì $\sqrt{x}-2\ge -2$

$\to \sqrt{x}\in\{3, 7, 1\}$

$\to x\in\{9, 49, 1\}$

e.Ta có:

$F=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}$ không có min, max