Đáp án:
b) \(\dfrac{1}{{x + 5}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne - 3;x \ne 5\\
\dfrac{{x + 1}}{{2\left( {x + 3} \right)}} + \dfrac{{x - 25}}{{5\left( {5 - x} \right)}}\\
= \dfrac{{5\left( {x + 1} \right)\left( {5 - x} \right) + 2\left( {x - 25} \right)\left( {x + 3} \right)}}{{10\left( {x + 3} \right)\left( {5 - x} \right)}}\\
= \dfrac{{5\left( {5x - {x^2} + 5 - x} \right) + 2\left( {{x^2} + 3x - 25x - 75} \right)}}{{10\left( {x + 3} \right)\left( {5 - x} \right)}}\\
= \dfrac{{ - 5{x^2} + 20x + 25 + 2{x^2} - 44x - 150}}{{10\left( {x + 3} \right)\left( {5 - x} \right)}}\\
= \dfrac{{ - 3{x^2} - 22x - 125}}{{10\left( {x + 3} \right)\left( {5 - x} \right)}}\\
b)DK:x \ne \left\{ { - 5;5} \right\}\\
\dfrac{{x - 5 + 2\left( {x + 5} \right) - 2x - 10}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}\\
= \dfrac{{ - x - 15 + 2x + 10}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}\\
= \dfrac{{x - 5}}{{\left( {x - 5} \right)\left( {x + 5} \right)}} = \dfrac{1}{{x + 5}}
\end{array}\)
