Giải bài 5.95 - 5.96 trang 215 - SBT Đại số và Giải tích lớp 11

5.95.

Ta có:

$\begin{aligned} & y=\dfrac{2x}{2\left( x-1 \right)\left( x+1 \right)}=\dfrac{x+1+x-1}{2\left( x-1 \right)\left( x+1 \right)} \\ & =\dfrac{1}{2\left( x-1 \right)}+\dfrac{1}{2\left( x+1 \right)}=\dfrac{1}{2}\left( \dfrac{1}{x-1}+\dfrac{1}{x+1} \right) \\ \end{aligned}$
Suy ra:

$\begin{aligned} & y'=\dfrac{1}{2}\left[ -\dfrac{1}{{{\left( x-1 \right)}^{2}}}-\dfrac{1}{{{\left( x+1 \right)}^{2}}} \right] \\ & y''=\dfrac{1}{2}\left[ \dfrac{2}{{{\left( x-1 \right)}^{3}}}+\dfrac{2}{{{\left( x+1 \right)}^{3}}} \right]=\dfrac{1}{{{\left( x-1 \right)}^{3}}}+\dfrac{1}{{{\left( x+1 \right)}^{3}}} \\ \end{aligned}$

5.96.

Ta có: 

$\begin{aligned} & y'=\dfrac{\left( x-2 \right)-\left( x+1 \right)}{{{\left( x-2 \right)}^{2}}}=\dfrac{-3}{{{\left( x-2 \right)}^{2}}} \\ & y''=\dfrac{6}{{{\left( x-2 \right)}^{3}}} \\ \end{aligned}$