Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
2x + 10 \ne 0\\
x \ne 0\\
2x\left( {x + 5} \right) \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ne 0\\
x \ne - 5
\end{array} \right.\\
b)x \ne 0;x \ne - 5\\
A = \frac{{{x^2} + 2x}}{{2x + 10}} + \frac{{x - 5}}{x} + \frac{{50 - 5x}}{{2x\left( {x + 5} \right)}}\\
= \frac{{\left( {{x^2} + 2x} \right)}}{{2\left( {x + 5} \right)}} + \frac{{x - 5}}{x} + \frac{{50 - 5x}}{{2x\left( {x + 5} \right)}}\\
= \frac{{\left( {{x^2} + 2x} \right).x + \left( {x - 5} \right).2\left( {x + 5} \right) + 50 - 5x}}{{2x\left( {x + 5} \right)}}\\
= \frac{{{x^3} + 2{x^2} + 2\left( {{x^2} - 25} \right) + 50 - 5x}}{{2x\left( {x + 5} \right)}}\\
= \frac{{{x^3} + 2{x^2} + 2{x^2} - 50 + 50 - 5x}}{{2x\left( {x + 5} \right)}}\\
= \frac{{{x^3} + 4{x^2} - 5x}}{{2x\left( {x + 5} \right)}}\\
= \frac{{x\left( {{x^2} + 4x - 5} \right)}}{{2x\left( {x + 5} \right)}}\\
= \frac{{\left( {x + 5} \right)\left( {x - 1} \right)}}{{2\left( {x + 5} \right)}}\\
= \frac{{x - 1}}{2}\\
A = 1 \Rightarrow \frac{{x - 1}}{2} = 1 \Rightarrow x = 3\left( {tmdk} \right)\\
A = - 3 \Rightarrow \frac{{x - 1}}{2} = - 3 \Rightarrow x = - 5\left( {ktm} \right)
\end{array}$
Vậy ko có x để A =-3 và x=3 thì A=1
