Exponentielle – Ex1

Exercice 1

Prouver, que pour tout $x \in \R$ :

  1. $\dfrac{1-\text{e}^{-2x}}{1+\text{e}^{-2x}}=\dfrac{\text{e}^{2x}-1}{\text{e}^{2x}+1}$
    $\quad$
  2. $\text{e}^{-x}-\text{e}^{-2x}=\dfrac{\text{e}^x-1}{\text{e}^{2x}}$
    $\quad$
  3. $\left(\text{e}^x+\text{e}^{-x}\right)^2-2=\dfrac{\text{e}^{4x}+1}{\text{e}^{2x}}$

 

Correction

 

  1. $\quad$
    $\begin{align}\dfrac{1-\text{e}^{-2x}}{1+\text{e}^{-2x}} &=\dfrac{\text{e}^{-2x}\left(\text{e}^{2x}-1\right)}{\text{e}^{-2x}\left(\text{e}^{2x} + 1\right)} \\\\
    &=\dfrac{\text{e}^{2x}-1}{\text{e}^{2x}+1}
    \end{align}$
    $\quad$
  2. $\quad$
    $\begin{align} \text{e}^{-x}-\text{e}^{-2x} &= \text{e}^{-2x}\left(\dfrac{ \text{e}^{-x}}{ \text{e}^{-2x}} – 1\right) \\\\
    &=\dfrac{1}{ \text{e}^{2x}}\left( \text{e}^{x} – 1\right)\\\\
    &=\dfrac{\text{e}^x-1}{\text{e}^{2x}}
    \end{align}$
    $\quad$
  3. $\quad$
    $\begin{align}
    \left(\text{e}^x+\text{e}^{-x}\right)^2-2&= \text{e}^{2x} + 2 +  \text{e}^{-2x} – 2\\\\
    &=  \text{e}^{2x}+ \text{e}^{-2x} \\\\
    &=  \text{e}^{-2x}\left(\dfrac{ \text{e}^{2x}}{ \text{e}^{-2x}} + 1\right)\\\\
    &=\dfrac{\text{e}^{4x}+1}{\text{e}^{2x}}
    \end{align}$