Đáp án:

b. \(\dfrac{{3{x^3} - 10{x^2} + 40x + 750}}{{2x\left( {x - 5} \right)\left( {x + 5} \right)}}\)

Giải thích các bước giải:

\(\begin{array}{l}
a.DK:x \ne \left\{ { - 5;0;5} \right\}\\
b.\dfrac{{{x^2} + 2x}}{{2\left( {x + 5} \right)}} + \dfrac{{x - 5}}{x} + \dfrac{{50 - x}}{{x\left( {x - 5} \right)}}\\
 = \dfrac{{{x^3} + 2{x^2} + 2\left( {{x^2} - 25} \right)\left( {x - 5} \right) + 2.\left( {50 - x} \right)\left( {x + 5} \right)}}{{2x\left( {x - 5} \right)\left( {x + 5} \right)}}\\
 = \dfrac{{{x^3} + 2{x^2} + 2\left( {{x^3} - 5{x^2} - 25x + 125} \right) + 2\left( {50x + 250 - {x^2} - 5x} \right)}}{{2x\left( {x - 5} \right)\left( {x + 5} \right)}}\\
 = \dfrac{{{x^3} + 2{x^2} + 2{x^3} - 10{x^2} - 50x + 250 + 100x + 500 - 2{x^2} - 10x}}{{2x\left( {x - 5} \right)\left( {x + 5} \right)}}\\
 = \dfrac{{3{x^3} - 10{x^2} + 40x + 750}}{{2x\left( {x - 5} \right)\left( {x + 5} \right)}}\\
c.\dfrac{{3{x^3} - 10{x^2} + 40x + 750}}{{2x\left( {x - 5} \right)\left( {x + 5} \right)}} = 1\\
 \to 3{x^3} - 10{x^2} + 40x + 750 = 2x\left( {{x^2} - 25} \right)\\
 \to 3{x^3} - 10{x^2} + 40x + 750 = 2{x^3} - 50x\\
 \to {x^3} - 10{x^2} + 90x + 750 = 0\\
 \to x =  - 4,709090214
\end{array}\)