Đáp án:
$a. P = \frac{x-1}{\sqrt[]{x}}$
$b. P = - 2$
Giải thích các bước giải:
ĐKXĐ : $x ≥ 0 , x \ne 1$
$a. P = ( \frac{\sqrt[]{x}}{\sqrt[]{x}-1} - \frac{1}{x-\sqrt[]{x}} ) : ( \frac{1}{1+\sqrt[]{x}} + \frac{2}{x-1} )$
⇔ $P = ( \frac{\sqrt[]{x}.\sqrt[]{x}}{\sqrt[]{x}(\sqrt[]{x}-1)} - \frac{1}{\sqrt[]{x}(\sqrt[]{x}-1)} ) : ( \frac{\sqrt[]{x}-1}{(\sqrt[]{x}-1)(\sqrt[]{x}+1)} + \frac{2}{(\sqrt[]{x}-1)(\sqrt[]{x}+1)} )$
⇔ $P = \frac{x-1}{\sqrt[]{x}(\sqrt[]{x}-1)} : \frac{\sqrt[]{x}+1}{(\sqrt[]{x}-1)(\sqrt[]{x}+1)}$
⇔ $P = \frac{(\sqrt[]{x}-1)(\sqrt[]{x}+1)}{\sqrt[]{x}(\sqrt[]{x}-1)} . \frac{(\sqrt[]{x}-1)(\sqrt[]{x}+1)}{\sqrt[]{x}+1}$
⇔ $P = \frac{(\sqrt[]{x}-1)(\sqrt[]{x}+1)}{\sqrt[]{x}}$
⇔ $P = \frac{x-1}{\sqrt[]{x}}$
$b. x = 3 - 2\sqrt[]{2}$
⇔ $x = 2 - 2\sqrt[]{2} + 1$
⇔ $x = ( \sqrt[]{2} - 1 )^{2}$
⇔ $\sqrt[]{x} = | \sqrt[]{2} - 1 |$
⇒ $\sqrt[]{x} = \sqrt[]{2} - 1$ ( vì $\sqrt[]{2} - 1 > 0$ )
Thay vào $P$ được :
$P = \frac{3-2\sqrt[]{2}-1}{\sqrt[]{2}-1}$
⇔ $P = \frac{2-2\sqrt[]{2}}{\sqrt[]{2}-1}$
⇔ $P = \frac{-2(\sqrt[]{2}-1)}{\sqrt[]{2}-1}$
⇔ $P = - 2$
