Lời giải:
a. ĐKXĐ: $x\geq 0; x\neq 1$
\(P=\frac{\sqrt{x}(\sqrt{x}+1)+3(\sqrt{x}-1)-(6\sqrt{x}-4)}{(\sqrt{x}-1)(\sqrt{x}+1)}\\ =\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{(\sqrt{x}-1)(\sqrt{x}+1)}\\ =\frac{x-2\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{(\sqrt{x}-1)^2}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b.
$x=4+2\sqrt{3}=(\sqrt{3}+1)^2\Rightarrow \sqrt{x}=\sqrt{3}+1$
Khi đó:
$P=\frac{\sqrt{3}+1-1}{\sqrt{3}+1+1}=-3+2\sqrt{3}$
c.
$P< \frac{1}{2}\Leftrightarrow \frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{1}{2}<0$
$\Leftrightarrow \frac{\sqrt{x}-3}{2(\sqrt{x}+1)}<0$
$\Leftrightarrow \sqrt{x}-3<0$
$\Leftrightarrow 0\leq x< 9$
Kết hợp đkxđ suy ra $0\leq x<9; x\neq 1$
a.
ĐKXĐ: \(x\ge0;x\ne1\)
\(P=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b.
\(x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}+1\)
\(\Rightarrow P=\dfrac{\sqrt{3}+1-1}{\sqrt{3}+1+1}=\dfrac{\sqrt{3}}{2+\sqrt{3}}=2\sqrt{3}-3\)
c.
\(P< \dfrac{1}{2}\Rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}< \dfrac{1}{2}\)
\(\Rightarrow2\left(\sqrt{x}-1\right)< \sqrt{x}+1\) (do \(\sqrt{x}+1>0\))
\(\Leftrightarrow\sqrt{x}< 3\)
\(\Leftrightarrow x< 9\)
kết hợp ĐKXD \(\Rightarrow\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)
Anh chị giúp em với ạ^^
