Ví dụ 9:
a.
$x-\sqrt{x}-\sqrt{xy}+\sqrt{y}=(x-\sqrt{xy})-(\sqrt{x}-\sqrt{y})$
$=\sqrt{x}(\sqrt{x}-\sqrt{y})-(\sqrt{x}-\sqrt{y})$
$=(\sqrt{x}-\sqrt{y})(\sqrt{x}-1)$
b.
$=\sqrt{(x-3)(x+3)}-2\sqrt{x-3}$
$=\sqrt{x-3}(\sqrt{x+3}-2)$
c.
$x-\sqrt{x}-6=(x+2\sqrt{x})-(3\sqrt{x}+6)$
$=\sqrt{x}(\sqrt{x}+2)-3(\sqrt{x}+2)=(\sqrt{x}+2)(\sqrt{x}-3)$
Bài 3:
ĐKXĐ: $x>0; x\neq 1$
a.
\(E=\frac{\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}-1)^2}:\left[\frac{(\sqrt{x}+1)(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}-1)}+\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}+\frac{2-x}{\sqrt{x}(\sqrt{x}-1)}\right]\)
\(=\frac{\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}-1)^2}:\frac{x-1+\sqrt{x}+2-x}{\sqrt{x}(\sqrt{x}-1)}\)
\(=\frac{\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}-1)^2}:\frac{\sqrt{x}+1}{\sqrt{x}(\sqrt{x}-1)}=\frac{\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}-1)^2}.\frac{\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}+1}=\frac{x}{\sqrt{x}-1}\)
b.
$E=\frac{x}{\sqrt{x}-1}>1$
$\Leftrightarrow \frac{x}{\sqrt{x}-1}-1>0$
$\Leftrightarrow \frac{x-\sqrt{x}+1}{\sqrt{x}-1}>0(*)$
Do $x-\sqrt{x}+1=(\sqrt{x}-\frac{1}{2})^2+\frac{3}{4}>0$ với mọi $x$ thuộc đkxđ nên $(*)\Leftrightarrow \sqrt{x}-1>0$
$\Leftrightarrow x>1$
Kết hợp đkxđ suy ra $x>1$
