Lời giải:
a. Khi $x=9$ thì:
$A=\frac{2\sqrt{9}+1}{\sqrt{9}}=\frac{7}{3}$
b.
\(B=\frac{x-3\sqrt{x}+4}{\sqrt{x}(\sqrt{x}-2)}-\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}-2)}=\frac{x-3\sqrt{x}+4-\sqrt{x}}{\sqrt{x}(\sqrt{x}-2)}=\frac{x-4\sqrt{x}+4}{\sqrt{x}(\sqrt{x}-2)}\\ =\frac{(\sqrt{x}-2)^2}{\sqrt{x}(\sqrt{x}-2)}=\frac{\sqrt{x}-2}{\sqrt{x}}\)
c.
\(P=B:A=\frac{\sqrt{x}-2}{\sqrt{x}}:\frac{2\sqrt{x}+1}{\sqrt{x}}=\frac{\sqrt{x}-2}{\sqrt{x}}.\frac{\sqrt{x}}{2\sqrt{x}+1}=\frac{\sqrt{x}-2}{2\sqrt{x}+1}\)
$|P|>P$
$\Leftrightarrow P<0$
$\Leftrightarrow \frac{\sqrt{x}-2}{2\sqrt{x}+1}<0$
$\Leftrightarrow \sqrt{x}-2<0$
$\Leftrightarrow 0\leq x<4$
Kết hợp với ĐKXĐ suy ra $0< x< 4$
a.
\(x=9\Rightarrow A=\dfrac{2\sqrt{9}+1}{\sqrt{9}}=\dfrac{7}{3}\)
b.
\(B=\dfrac{x-3\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
c.
\(P=\dfrac{B}{A}=\dfrac{\sqrt{x}-2}{\sqrt{x}}.\dfrac{\sqrt{x}}{2\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{2\sqrt{x}+1}\)
\(\left|P\right|>P\Leftrightarrow P< 0\)
\(\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}+1}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\) (do \(2\sqrt{x}+1>0;\forall x\))
\(\Leftrightarrow\sqrt{x}< 2\)
\(\Rightarrow x< 4\)
Kết hợp ĐKXĐ \(\Rightarrow0< x< 4\)
Anh chị giúp em với ạ^^
