Bài 2
$Đkx\ge0;x\ne4$
$Q=\bigg(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}\bigg):\dfrac{3\sqrt{x}-x}{x+4\sqrt{x}+4}\\=\dfrac{(\sqrt{x}+1)(\sqrt{x}+2)-2\sqrt{x}(\sqrt{x}-2)-5\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}.\dfrac{(\sqrt{x}+2)^2}{-\sqrt{x}(\sqrt{x}-3)}\\=\dfrac{x+3\sqrt{x}+2-2x+4\sqrt{x}-5\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}.\dfrac{(\sqrt{x}+2)^2}{-\sqrt{x}(\sqrt{x}-3)}\\=\dfrac{-\sqrt{x}(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}.\dfrac{(\sqrt{x}+2)^2}{-\sqrt{x}(\sqrt{x}-3)}\\=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}$
$Q=2\Leftrightarrow \dfrac{\sqrt{x}+2}{\sqrt{x}-3}=2\\\Leftrightarrow \sqrt{x}+2=2\sqrt{x}-6\\\Leftrightarrow \sqrt{x}=8\Leftrightarrow x=64\ (t/m)$
Vậy với $x=64$ thì $Q=2$
Bài 3
$Đk:a\ge0;a\ne9$
$B=\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3}{\sqrt{a}+3}-\dfrac{a-2}{a-9}\\=\dfrac{\sqrt{a}(\sqrt{a}+3)-3(\sqrt{a}-3)-a+2}{(\sqrt{a}-3)(\sqrt{a}+3)}\\=\dfrac{a+3\sqrt{a}-3\sqrt{a}+9-a+2}{a-9}\\=\dfrac{11}{a-9}$
$B$ nhận giá trị nguyên khi
$a-9=\{-11;-1;1;11\}\Leftrightarrow a=\{-2;8;10;20\}$
Kết hợp với điều kiện thì
Với $a=\{8;10;20\}$ thì $B$ nhận giá trị nguyên.
