Đáp án:

 a) x=5

Giải thích các bước giải:

\(\begin{array}{l}
a)DK:x \ne \left\{ { - 5;0} \right\}\\
A = \dfrac{{{x^2}}}{{5\left( {x + 5} \right)}} + \dfrac{{2\left( {x - 5} \right)}}{x} + \dfrac{{50 + 5x}}{{x\left( {x + 5} \right)}}\\
 = \dfrac{{{x^3} + 10\left( {{x^2} - 25} \right) + 5\left( {50 + 5x} \right)}}{{5x\left( {x + 5} \right)}}\\
 = \dfrac{{{x^3} + 10{x^2} - 250 + 250 + 25x}}{{5x\left( {x + 5} \right)}}\\
 = \dfrac{{{x^3} + 10{x^2} + 25x}}{{5x\left( {x + 5} \right)}}\\
 = \dfrac{{x{{\left( {x + 5} \right)}^2}}}{{5x\left( {x + 5} \right)}}\\
 = \dfrac{{x + 5}}{5}\\
A = 2\\
 \Leftrightarrow \dfrac{{x + 5}}{5} = 2\\
 \Leftrightarrow x + 5 = 10\\
 \Leftrightarrow x = 5\\
b)\dfrac{1}{A} = \dfrac{5}{{x + 5}}\\
\dfrac{1}{A} \in Z\\
 \Leftrightarrow \dfrac{5}{{x + 5}} \in Z\\
 \Leftrightarrow x + 5 \in U\left( 5 \right)\\
 \to \left[ \begin{array}{l}
x + 5 = 5\\
x + 5 =  - 5\\
x + 5 = 1\\
x + 5 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 0\\
x =  - 10\\
x =  - 4\\
x =  - 6
\end{array} \right.
\end{array}\)

Đáp án+Giải thích các bước giải:

$\begin{array}{l} a)DK:x \ne \left\{ { - 5;0} \right\}\\ A = \dfrac{{{x^2}}}{{5\left( {x + 5} \right)}} + \dfrac{{2\left( {x - 5} \right)}}{x} + \dfrac{{50 + 5x}}{{x\left( {x + 5} \right)}}\\ = \dfrac{{{x^3} + 10\left( {{x^2} - 25} \right) + 5\left( {50 + 5x} \right)}}{{5x\left( {x + 5} \right)}}\\ = \dfrac{{{x^3} + 10{x^2} - 250 + 250 + 25x}}{{5x\left( {x + 5} \right)}}\\ = \dfrac{{{x^3} + 10{x^2} + 25x}}{{5x\left( {x + 5} \right)}}\\ = \dfrac{{x{{\left( {x + 5} \right)}^2}}}{{5x\left( {x + 5} \right)}}\\ = \dfrac{{x + 5}}{5}\\ A = 2\\ \Leftrightarrow \dfrac{{x + 5}}{5} = 2\\ \Leftrightarrow x + 5 = 10\\ \Leftrightarrow x = 5\\ b)\dfrac{1}{A} = \dfrac{5}{{x + 5}}\\ \dfrac{1}{A} \in Z\\ \Leftrightarrow \dfrac{5}{{x + 5}} \in Z\\ \Leftrightarrow x + 5 \in U\left( 5 \right)\\ \to \left[ \begin{array}{l} x + 5 = 5\\ x + 5 = - 5\\ x + 5 = 1\\ x + 5 = - 1 \end{array} \right. \to \left[ \begin{array}{l} x = 0\\ x = - 10\\ x = - 4\\ x = - 6 \end{array} \right. \end{array}$