[tex]\dpi{150} A = \left ( 1-\frac{\sqrt{x}}{\sqrt{x}+1} \right ):\left ( \frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x+5\sqrt{x}+6} \right )[/tex]
a) Rút gọn A
b) Tìm [tex]\dpi{120} x \epsilon Z[/tex] để [tex]\dpi{120} A \epsilon Z[/tex]
c)Tìm x để A<0
d) Tìm x để [tex]\dpi{120} A \epsilon Z[/tex]
phải là $\left ( 1-\dfrac{\sqrt{x}}{\sqrt{x}+1} \right ):\left ( \dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x\color{red}{-}5\sqrt{x}+6} \right )$ chứ nhỉ?
a) ĐK: $x\geq 0;x\neq 4;x\neq 9$
$A=\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}-3)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{(\sqrt{x}-2)(\sqrt{x}-3)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}$
b) $A=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1-3}{\sqrt{x}+1}=1-\dfrac{3}{\sqrt{x}+1}$
$A\in \mathbb{Z}\Leftrightarrow \dfrac{3}{\sqrt{x}+1}\in \mathbb{Z}\Leftrightarrow (\sqrt{x}+1)\in Ư(3)$
$\Rightarrow \sqrt{x}+1\in \left \{ 1;3 \right \}\Rightarrow x\in \left \{ 0;4 \right \}$
Mà $x\neq 4\Rightarrow x=0$
c) $A<0\Leftrightarrow \dfrac{\sqrt{x}-2}{\sqrt{x}+1}<0\Leftrightarrow \sqrt{x}-2<0\Leftrightarrow \sqrt{x}<2\Leftrightarrow 0\leq x<4$
d) $A=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=1-\dfrac{3}{\sqrt{x}+1}\Rightarrow A\in \mathbb{Z}\Leftrightarrow \dfrac{3}{\sqrt x+1}\in \mathbb{Z}$
Mà $0<\dfrac{3}{\sqrt{x}+1}\leq 3\Rightarrow \dfrac{3}{\sqrt{x}+1}\in \left \{ 1;2;3 \right \}$
$+\dfrac{3}{\sqrt{x}+1}=1\Leftrightarrow x=4$ (L)
$+\dfrac{3}{\sqrt{x}+1}=2\Leftrightarrow x=\dfrac{1}{4}$ (N)
$+\dfrac{3}{\sqrt{x}+1}=3\Leftrightarrow x=0$ (N)
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
