Giúp với ạaaaa :)

$\text{a) Q = ($\dfrac{1}{\sqrt{x} + 1}$ - $\dfrac{2\sqrt{x} - 2}{x\sqrt{x} - \sqrt{x} + x - 1}$ ) : ($\dfrac{1}{\sqrt{x} - 1}$ - $\dfrac{2}{x - 1}$) Đkxđ: x $\geq$ 0; x $\neq$ 1}$

$\text{= [$\dfrac{1}{\sqrt{x} + 1}$ - $\dfrac{2(\sqrt{x} - 1)}{\sqrt{x}(x - 1) + (x - 1)}$ ] : $\dfrac{1(\sqrt{x} + 1) - 2}{(\sqrt{x} - 1)(\sqrt{x} + 1)}$ }$

$\text{= [$\dfrac{1}{\sqrt{x} + 1}$ - $\dfrac{2(\sqrt{x} - 1)}{(\sqrt{x} + 1)(x - 1)}$ ] . $\dfrac{(\sqrt{x} - 1)(\sqrt{x} + 1)}{\sqrt{x} + 1 - 2}$ }$

$\text{= [$\dfrac{1}{\sqrt{x} + 1}$ - $\dfrac{2(\sqrt{x} - 1)}{(\sqrt{x} + 1)(\sqrt{x} + 1)(\sqrt{x} - 1)}$ ] . $\dfrac{(\sqrt{x} - 1)(\sqrt{x} + 1)}{\sqrt{x} - 1}$}$

$\text{= [$\dfrac{1}{\sqrt{x} + 1}$ - $\dfrac{2}{(\sqrt{x} + 1)^2}$ ] . ($\sqrt{x}$ + 1)}$

$\text{= $\dfrac{1(\sqrt{x} + 1) - 2}{(\sqrt{x} + 1)^2}$ . ($\sqrt{x}$ + 1)}$

$\text{= $\dfrac{\sqrt{x} + 1 - 2}{\sqrt{x} + 1}$ }$

$\text{= $\dfrac{\sqrt{x} - 1}{\sqrt{x} + 1}$ }$

$\text{b) Q = $\dfrac{\sqrt{x} - 1}{\sqrt{x} + 1}$ Đkxđ: x ≥ 0; x $\neq$ 1}$

$\text{= 1 - $\dfrac{2}{\sqrt{x} + 1}$ }$

$\text{Có: $\sqrt{x}$ ≥ 0 với ∀ x ∈ Đkxđ}$

$\text{⇔ $\sqrt{x}$ + 1 ≥ 1 }$

$\text{⇔ $\dfrac{2}{\sqrt{x} + 1}$ ≤ 2}$

$\text{⇔ -$\dfrac{2}{\sqrt{x} + 1}$ ≥ -2}$

$\text{⇔ 1 - $\dfrac{2}{\sqrt{x} + 1}$ ≥ -1}$

$\text{⇔ Q ≥ -1}$

$\text{Dấu "=" xảy ra ⇔ x = 0 (t/m)}$

$\text{Vậy $Q_{min}$ = -1 khi x = 0}$

$\textit{Ha1zzz2511}$