
\begin{enumerate}
\item $\begin{aligned}[t]
(3x-5)^2=16 & \Longleftrightarrow \left(
\begin{array}{l}
3x-5=4\\
\kern-.75em\text{ou}\\
3x-5=-4
\end{array}
\right.
\\
&
\Longleftrightarrow \left(
\begin{array}{l}
3x=9\\
\kern-.75em\text{ou}\\
3x=1
\end{array}
\right.
\\
&
\Longleftrightarrow \left(
\begin{array}{l}
x=3\\
\kern-.75em\text{ou}\\
x=\dfrac{1}{3}
\end{array}
\right.
\end{aligned}$

On a alors : $\boxed{\mathcal{S}=\left\{3~;\dfrac{1}{3}\right\}}$.

\item $\begin{aligned}[t]
\big(x^2-4\big)^2=81 &
\Longleftrightarrow \left(
\begin{array}{l}
x^2-4=9\\
\kern-.75em\text{ou}\\
x^2-4=-9
\end{array}
\right.
\\
&
\Longleftrightarrow \left(
\begin{array}{l}
x^2=13\\
\kern-.75em\text{ou}\\
x^2=-5~~\longleftarrow\text{impossible car }x^2\geqslant0
\end{array}
\right.
\\
&
\Longleftrightarrow \left(
\begin{array}{l}
x=\sqrt{13}\\
\kern-.75em\text{ou}\\
x=-\sqrt{13}
\end{array}
\right.
\\
\end{aligned}$

On a alors : $\boxed{\mathcal{S}=\big\{-\sqrt{13}~;\sqrt{13}\big\}}$.
\newpage
\item $\begin{aligned}[t]
\big(3x^2-4\big)^2=9 & \Longleftrightarrow
\left(
\begin{array}{l}
3x^2-4=3\\
\kern-.75em\text{ou}\\
3x^2-4=-3
\end{array}
\right.
\\
&
\Longleftrightarrow
\left(
\begin{array}{l}
3x^2=7\\
\kern-.75em\text{ou}\\
3x^2=1
\end{array}
\right.
\\
&
 \Longleftrightarrow
\left(
\begin{array}{l}
x^2=\dfrac{7}{3}\\
\kern-.75em\text{ou}\\
x^2=\dfrac{1}{3}
\end{array}
\right.
\\
&
 \Longleftrightarrow
\left(
\begin{array}{l}
x=\sqrt{\dfrac{7}{3}}\qquad\text{ou}\qquad x=-\sqrt{\dfrac{7}{3}}\\
\kern-.75em\text{ou}\\
x=\dfrac{1}{\sqrt{3}}\qquad\text{ou}\qquad x=-\dfrac{1}{\sqrt{3}}
\end{array}
\right.
\end{aligned}$

On a alors : $\boxed{\mathcal{S}=\left\{\sqrt{\dfrac{7}{3}}~;-\sqrt{\dfrac{7}{3}}~;\dfrac{1}{\sqrt{3}}~;-\dfrac{1}{\sqrt{3}} \right\}}$.

\item $\begin{aligned}[t]
\left(\dfrac{2x-1}{x-7}\right)^2=25
&
\Longleftrightarrow
\left\{
	\begin{array}{l}
		\left(
			\begin{array}{l}
				\dfrac{2x-1}{x-7}=5\\
				\kern-.75em\text{ou}\\
				\dfrac{2x-1}{x-7}=-5
			\end{array}
		\right.
		\\
		\kern2em x-7\neq0
	\end{array}
\right.\\
&
\Longleftrightarrow
\left\{
	\begin{array}{l}
		\left(
			\begin{array}{l}
				\dfrac{2x-1}{x-7}-5=0\\
				\kern-.75em\text{ou}\\
				\dfrac{2x-1}{x-7}+5=0
			\end{array}
		\right.
		\\
		\kern2em x\neq7
	\end{array}
\right.\\
&
\Longleftrightarrow
\left\{
	\begin{array}{l}
		\left(
			\begin{array}{l}
				\dfrac{2x-1}{x-7}-\dfrac{5(x-7)}{x-7}=0\\
				\kern-.75em\text{ou}\\
				\dfrac{2x-1}{x-7}+\dfrac{5(x-7)}{x-7}=0
			\end{array}
		\right.
		\\
		\kern2em x\neq7
	\end{array}
\right.\\
&
\Longleftrightarrow
\left\{
	\begin{array}{l}
		\left(
			\begin{array}{l}
				\dfrac{2x-1}{x-7}-\dfrac{5x-35}{x-7}=0\\
				\kern-.75em\text{ou}\\
				\dfrac{2x-1}{x-7}+\dfrac{5x-35}{x-7}=0
			\end{array}
		\right.
		\\
		\kern2em x\neq7
	\end{array}
\right.\\
&
\Longleftrightarrow
\left\{
	\begin{array}{l}
		\left(
			\begin{array}{l}
				\dfrac{2x-1-5x+35}{x-7}=0\\
				\kern-.75em\text{ou}\\
				\dfrac{2x-1+5x-35}{x-7}=0
			\end{array}
		\right.
		\\
		\kern2em x\neq7
	\end{array}
\right.\\
\end{aligned}$

$\begin{aligned}[t]
\left(\dfrac{2x-1}{x-7}\right)^2=25
&
\Longleftrightarrow
\left\{
	\begin{array}{l}
		\left(
			\begin{array}{l}
				2x-1-5x+35=0\\
				\kern-.75em\text{ou}\\
				2x-1+5x-35=0
			\end{array}
		\right.
		\\
		\kern2em x\neq7
	\end{array}
\right.\\
&
\Longleftrightarrow
\left\{
	\begin{array}{l}
		\left(
			\begin{array}{l}
				-3x+34=0\\
				\kern-.75em\text{ou}\\
				7x-36=0
			\end{array}
		\right.
		\\
		\kern2em x\neq7
	\end{array}
\right.\\
&
\Longleftrightarrow
\left\{
	\begin{array}{l}
		\left(
			\begin{array}{l}
				x=\dfrac{34}{3}\\
				\kern-.75em\text{ou}\\
				x=\dfrac{36}{7}
			\end{array}
		\right.
		\\[30pt]
		\kern2em x\neq7
	\end{array}
\right.
\end{aligned}$

\bigskip

Ainsi, $\boxed{\mathcal{S}=\left\{\dfrac{36}{7}~;\dfrac{34}{3} \right\}}$.
\end{enumerate}
