Cho biểu thức P =
a) tìm ĐKXĐ
b) Rút gọn P
c) tìm x để P>3
d) Tìm x
Q để P
Z
a) ĐKXĐ: $x\geq 0;x\neq 4$
b) $P=\dfrac{\sqrt x+1}{\sqrt x-2}+\dfrac{2\sqrt x}{\sqrt x+2}+\dfrac{2+5\sqrt x}{4-x}
\\=\dfrac{(\sqrt x+1)(\sqrt x+2)+2\sqrt x(\sqrt x-2)-(2+5\sqrt x)}{(\sqrt x-2)(\sqrt x+2)}
\\=\dfrac{x+3\sqrt x+2+2x-4\sqrt x-2-5\sqrt x}{(\sqrt x-2)(\sqrt x+2)}
\\=\dfrac{3x-6\sqrt x}{(\sqrt x-2)(\sqrt x+2)}=\dfrac{3\sqrt x(\sqrt x-2)}{(\sqrt x-2)(\sqrt x+2)}
\\=\dfrac{3\sqrt x}{\sqrt x+2}$
c) $P>3\Leftrightarrow \dfrac{3\sqrt x}{\sqrt x+2}-3>0\Leftrightarrow \dfrac{-6}{\sqrt x+2}>0\Leftrightarrow \sqrt{x}+2<0\Leftrightarrow \sqrt{x}<-2$ (vô lí)
Vậy ko có giá trị nào của $x$ để $P>3$ or là đề sai ^^.
d) $P=\dfrac{3\sqrt{x}}{\sqrt{x}+2}=3-\dfrac{6}{\sqrt{x}+2}\Rightarrow P\in \mathbb{Z}\Leftrightarrow \dfrac{6}{\sqrt{x}+2}\in \mathbb{Z}\Leftrightarrow \sqrt x+2\in Ư(6)$
Mà $\sqrt{x}+2\geq 2\Rightarrow \sqrt{x}+2\in \left \{ 2;3;6 \right \}\Rightarrow x\in \left \{ 0;1;16 \right \}$ (TM)
Vậy...