Giải bài 5.18 trang 202 - SBT Đại số và Giải tích lớp 11

$\begin{align} f\left( x \right)&=\left[ \dfrac{x-1}{2(\sqrt{x}+1)}+1 \right].\dfrac{2}{\sqrt{x}+1}:{{\left( \dfrac{\sqrt{x-2}}{\sqrt{x+2}+\sqrt{x-2}}+\dfrac{x-2}{\sqrt{{{x}^{2}}-4}-x+2} \right)}^{2}} \\ & =\dfrac{x+2\sqrt{x}+1}{2\left( \sqrt{x}+1 \right)}.\dfrac{2}{\sqrt{x}+1}:{{\left[ \dfrac{\sqrt{x-2}}{\sqrt{x+2}+\sqrt{x-2}}+\dfrac{{{\left( \sqrt{x-2} \right)}^{2}}}{\sqrt{x-2}\left( \sqrt{x+2}-\sqrt{x-2} \right)} \right]}^{2}} \\ & =1:{{\left( \dfrac{\sqrt{x-2}}{\sqrt{x+2}+\sqrt{x-2}}+\dfrac{\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-2}} \right)}^{2}} \\ & =1:{{\left[ \dfrac{\sqrt{{{x}^{2}}-4}-\left( x-2 \right)+\left( x-2 \right)+\sqrt{{{x}^{2}}-4}}{4} \right]}^{2}} \\ & =1:{{\left( \dfrac{\sqrt{{{x}^{2}}-4}}{2} \right)}^{2}}=\dfrac{4}{{{x}^{2}}-4} \\ \end{align}$ 
Suy ra $f'\left( x \right)=\left( \dfrac{4}{{{x}^{2}}-4} \right)'=-\dfrac{4\left( 2x \right)}{{{\left( {{x}^{2}}-4 \right)}^{2}}}=-\dfrac{8x}{{{\left( {{x}^{2}}-4 \right)}^{2}}}$