Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
2x + 10 \ne 0\\
x \ne 0\\
2x\left( {x + 5} \right) \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ne - 5\\
x \ne 0
\end{array} \right.\\
b)x \ne - 5;x \ne 0\\
B = \dfrac{{{x^2} + 2x}}{{2x + 10}} + \dfrac{{x - 5}}{x} + \dfrac{{50 - 5x}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{\left( {{x^2} + 2x} \right).x + 2\left( {x + 5} \right)\left( {x - 5} \right) + 50 - 5x}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{{x^3} + 2{x^2} + 2{x^2} - 50 + 50 - 5x}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{{x^3} + 4{x^2} - 5x}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{x\left( {{x^2} + 4x - 5} \right)}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{x\left( {x + 5} \right)\left( {x - 1} \right)}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{x - 1}}{2}\\
B = 0\\
\Rightarrow \dfrac{{x - 1}}{2} = 0 \Rightarrow x = 1\left( {tmdk} \right)\\
B = \dfrac{1}{4}\\
\Rightarrow \dfrac{{x - 1}}{2} = \dfrac{1}{4}\\
\Rightarrow x - 1 = \dfrac{1}{2}\\
\Rightarrow x = \dfrac{3}{2}\left( {tmdk} \right)
\end{array}$
