Cứu mik vs càng nhanh càng tốt nha

Đáp án:

\(\begin{array}{l}
1)x = \dfrac{3}{{20}}\\
2)x = 1\\
3)x \in \emptyset 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
1)DK:x \ne  \pm 5\\
\dfrac{{x + 5}}{{x - 5}} - \dfrac{{x - 5}}{{x + 5}} = \dfrac{3}{{{x^2} - 25}}\\
 \to \dfrac{{{{\left( {x + 5} \right)}^2} - {{\left( {x - 5} \right)}^2} - 3}}{{\left( {x - 5} \right)\left( {x + 5} \right)}} = 0\\
 \to {x^2} + 10x + 25 - {x^2} + 10x - 25 - 3 = 0\\
 \to 20x - 3 = 0\\
 \to x = \dfrac{3}{{20}}\\
2)DK:x \ne \left\{ {0;2} \right\}\\
\dfrac{{x + 2}}{{x - 2}} - \dfrac{5}{x} = \dfrac{8}{{{x^2} - 2x}}\\
 \to \dfrac{{x\left( {x + 2} \right) - 5\left( {x - 2} \right) - 8}}{{x\left( {x - 2} \right)}} = 0\\
 \to \dfrac{{{x^2} + 2x - 5x + 10 - 8}}{{x\left( {x - 2} \right)}} = 0\\
 \to {x^2} - 3x + 2 = 0\\
 \to \left( {x - 1} \right)\left( {x - 2} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 1\\
x = 2\left( l \right)
\end{array} \right.\\
3)DK:x \ne 2\\
\dfrac{1}{{x - 2}} - \dfrac{{x - 3}}{{2 - x}} =  - 3\\
 \to 1 + x - 3 =  - 3\left( {x - 2} \right)\\
 \to x - 2 =  - 3x + 6\\
 \to 4x = 8\\
 \to x = 2\left( l \right)\\
 \to x \in \emptyset 
\end{array}\)