$a,$ $(x-2)(2x-1)=5(2-x)$

$⇔2x^2-x-4x+2=10-5x$

$⇔2x^2-5x+2=10-5x$

$⇔2x^2-5x+5x-10+2=0$

$⇔2x^2-8=0$

$⇔2(x^2-4)=0$

$⇔x^2=4$

$⇔x=±2$

Vậy phương trình có nghiệm `S={±2}`

$b,$ $ĐKXĐ:x\neq ±5$

$\dfrac{x+5}{x-5}-\dfrac{x-5}{x+5}=\dfrac{x(x+25)}{x^2-25}$ 

$⇔\dfrac{x^2+10x+25}{(x-5)(x+5)}-\dfrac{x^2-10x+25}{(x+5)(x-5)}=\dfrac{x(x+25)}{(x-5)(x+5)}$ 

$⇔\dfrac{x^2+10x+25-x^2+10x-25}{(x-5)(x+5)}-\dfrac{x^2-10x+25}{(x+5)(x-5)}=\dfrac{x(x+25)}{(x-5)(x+5)}$ 

$⇔20x=x(x+25)$

$⇔x^2+25x-20x=0$

$⇔x^2+5x=0$

$⇔$ \(\left[ \begin{array}{l}x=0\\x+5=0⇒x=-5\end{array} \right.\) 

Vậy phương trình có nghiệm `S={0;-5}`.

#andy

\[\begin{array}{l}
\dfrac{{x + 5}}{{x - 5}} - \dfrac{{x - 5}}{{x + 5}} = \dfrac{{x(x + 25)}}{{{x^2} - 25}}\\
DKXD\left\{ \begin{array}{l}
x - 5 \ne 0\\
x + 5 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne 5\\
x \ne  - 5
\end{array} \right.\\
 \Leftrightarrow \dfrac{{x + 5}}{{x - 5}} - \dfrac{{x - 5}}{{x + 5}} = \dfrac{{x\left( {x + 25} \right)}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}\\
 \Leftrightarrow \dfrac{{\left( {x + 5} \right)\left( {x + 5} \right)}}{{\left( {x - 5} \right)\left( {x + 5} \right)}} - \dfrac{{\left( {x - 5} \right)\left( {x - 5} \right)}}{{\left( {x - 5} \right)\left( {x + 5} \right)}} = \dfrac{{x\left( {x + 25} \right)}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}\\
 \Leftrightarrow \dfrac{{{{\left( {x + 5} \right)}^2}}}{{\left( {x - 5} \right)\left( {x + 5} \right)}} - \dfrac{{{{\left( {x - 5} \right)}^2}}}{{\left( {x - 5} \right)\left( {x + 5} \right)}} = \dfrac{{x\left( {x + 25} \right)}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}\\
 \Leftrightarrow \dfrac{{{x^2} + 10x + 25}}{{(x - 5)(x + 5)}} - \dfrac{{{x^2} - 10x + 25}}{{(x + 5)(x - 5)}} = \dfrac{{x(x + 25)}}{{(x - 5)(x + 5)}}\\
 \Leftrightarrow \dfrac{{{x^2} + 10x + 25 - {x^2} + 10x - 25}}{{\left( {x - 5} \right)\left( {x + 5} \right)}} = \dfrac{{x\left( {x + 25} \right)}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}\\
 \Leftrightarrow \dfrac{{20x}}{{\left( {x - 5} \right)\left( {x + 5} \right)}} = \dfrac{{x\left( {x + 25} \right)}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}\\
 \Leftrightarrow 20x = x\left( {x + 25} \right)\\
 \Leftrightarrow 20x = {x^2} + 25x\\
 \Leftrightarrow 20x - {x^2} - 25x = 0\\
 \Leftrightarrow  - 5x - {x^2} = 0\\
 \Leftrightarrow  - x\left( {5 + x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
 - x = 0\\
5 + x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\left( {TMDK} \right)\\
x =  - 5\left( {KTMDK} \right)
\end{array} \right.\\
 \Rightarrow S \in \left\{ 0 \right\}
\end{array}\]