a. \(P=\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{3+\sqrt{x}}-\dfrac{9-x}{x+\sqrt{x}-6}\)
\(=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{3+\sqrt{x}}-\dfrac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\left(9-x\right)+\left(\sqrt{x}-2\right)^2-\left(9-x\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\) \(=\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\)
b. Để \(P=\dfrac{7}{12}\) thì \(\dfrac{\sqrt{x}-2}{\sqrt{x}+3}=\dfrac{7}{12}\)
\(\Rightarrow12\left(\sqrt{x}-2\right)=7\left(\sqrt{x}+3\right)\)
\(\Rightarrow12\sqrt{x}-7\sqrt{x}=24+21\)
\(\Rightarrow5\sqrt{x}=45\) \(\Rightarrow\sqrt{x}=9\) \(\Rightarrow x=81\)
Vậy để \(P=\dfrac{7}{12}\) thì \(x=81\)
c. Để \(P>\dfrac{1}{2}\) thì \(\dfrac{\sqrt{x}-2}{\sqrt{x}+3}>\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+3}-\dfrac{1}{2}>0\) \(\Leftrightarrow\dfrac{2\sqrt{x}-4-\sqrt{x}-3}{2\sqrt{x}+6}>0\) \(\Leftrightarrow\dfrac{\sqrt{x}-7}{2\sqrt{x}+6}>0\) (*)
Mà \(x\ge0\Leftrightarrow\sqrt{x}\ge0\Leftrightarrow2\sqrt{x}+6\ge6>0\)
(*) \(\Rightarrow\sqrt{x}-7>0\Rightarrow\sqrt{x}>7\Rightarrow x>49\)
Vậy để \(P>\dfrac{1}{2}\) thì \(x>49\)
d. Ta có \(\dfrac{1}{P}=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}=\dfrac{\sqrt{x}-2+5}{\sqrt{x}-2}=1+\dfrac{5}{\sqrt{x}-2}\)
Để \(\dfrac{1}{P}\) nhận giá trị nguyên thì \(5⋮\left(\sqrt{x}-2\right)\Rightarrow\left(\sqrt{x}-2\right)\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}-2=1\\\sqrt{x}-2=-1\\\sqrt{x}-2=5\\\sqrt{x}-2=-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=1\\\sqrt{x}=7\\\sqrt{x}=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=9\\x=1\\x=49\end{matrix}\right.\)
Vậy tất cả các x thỏa mãn ycbt là x=9, x=1 hoặc x=49
e. Ta có \(P=\dfrac{\sqrt{x}-2}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3-5}{\sqrt{x}+3}=1-\dfrac{5}{\sqrt{x}+3}\)
Để \(P\) nhận giá trị nguyên thì \(5⋮\left(\sqrt{x}+3\right)\Rightarrow\left(\sqrt{x}+3\right)\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+3=1\\\sqrt{x}+3=-1\\\sqrt{x}+3=5\\\sqrt{x}+3=-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=-2\\\sqrt{x}=-4\\\sqrt{x}=2\\\sqrt{x}=-8\end{matrix}\right.\) \(\Rightarrow x=4\)
Vậy x thỏa mãn ycbt là x=4
Làm nốt câu e:
\(P=\frac{\sqrt{x}-2}{\sqrt{x}+3}=1-\frac{5}{\sqrt{x}+3}\)
Với mọi $x\geq 0; x\neq 4$ thì $\frac{5}{\sqrt{x}+3}>0$
$\Rightarrow P< 1$
Mặt khác: $\sqrt{x}+3\geq 3$
$\Rightarrow \frac{5}{\sqrt{x}+3}\leq \frac{5}{3}$
$\Rightarrow P\geq 1-\frac{5}{3}=\frac{-2}{3}$
Vậy $1> P\geq \frac{-2}{3}$
$P$ nguyên $\Leftrightarrow P=0$
$\Leftrightarrow \frac{\sqrt{x}-2}{\sqrt{x}+3}=0$
$\Leftrightarrow x=4$ (trái đkxđ)
Vậy không tồn tại $x$ để $P$ nguyên.
a: Ta có: \(P=\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}-\dfrac{9-x}{x+\sqrt{x}-6}\)
\(=\dfrac{-x+9+x-4\sqrt{x}+4-9+x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-4\sqrt{x}+4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\)
b: Để \(P=\dfrac{7}{12}\) thì \(12\sqrt{x}-24=7\sqrt{x}+21\)
\(\Leftrightarrow5\sqrt{x}=45\)
\(\Leftrightarrow x=81\)
