Giải thích các bước giải:
a,
ĐKXĐ:
\(\left\{ \begin{array}{l}
x \ne 0\\
x + 5 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne 0\\
x \ne - 5
\end{array} \right.\)
b,
Ta có:
\(\begin{array}{l}
A = \dfrac{{{x^2} + 2x}}{{2x + 10}} + \dfrac{{x - 5}}{x} + \dfrac{{50 - 5x}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{{x^2} + 2x}}{{2\left( {x + 5} \right)}} + \dfrac{{x - 5}}{x} + \dfrac{{50 - 5x}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{\left( {{x^2} + 2x} \right).x}}{{2x\left( {x + 5} \right)}} + \dfrac{{2\left( {x - 5} \right)\left( {x + 5} \right)}}{{2x\left( {x + 5} \right)}} + \dfrac{{50 - 5x}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{{x^3} + 2{x^2} + 2\left( {{x^2} - 25} \right) + 50 - 5x}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{{x^3} + 2{x^2} + 2{x^2} - 50 + 50 - 5x}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{{x^3} + 4{x^2} - 5x}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{x\left( {{x^2} + 4x - 5} \right)}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{x\left( {x + 5} \right)\left( {x - 1} \right)}}{{2x\left( {x + 5} \right)}} = \dfrac{{x - 1}}{2}
\end{array}\)
\(\begin{array}{l}
A = 1 \Leftrightarrow \dfrac{{x - 1}}{2} = 1 \Leftrightarrow x - 1 = 2 \Leftrightarrow x = 3\\
A = - 3 \Leftrightarrow \dfrac{{x - 1}}{2} = - 3 \Leftrightarrow x - 1 = - 6 \Leftrightarrow x = - 5
\end{array}\)
