Đáp án:

\(\begin{array}{l}
a.M = 1 + \frac{5}{x}\\
b.x =  \pm 10
\end{array}\) 

c. x∈{-5,-1,1,5}

Giải thích các bước giải:

\(\begin{array}{l}
a.M = \left( {\frac{x}{{x + 5}} - \frac{5}{{5 - x}} + \frac{{10x}}{{{x^2} - 25}}} \right).\left( {1 - \frac{5}{x}} \right)(dk:x \ne 0,x \ne  \pm 5)\\
 = \left[ {\frac{{x(x - 5)}}{{{x^2} - 25}} + \frac{{5(x + 5)}}{{{x^2} - 25}} + \frac{{10x}}{{{x^2} - 25}}} \right].\frac{{x - 5}}{x}\\
 = \frac{{{x^2} - 5x + 5x + 25 + 10x}}{{(x - 5)(x + 5)}}.\frac{{x - 5}}{x}\\
 = \frac{{{{(x + 5)}^2}}}{{(x - 5)(x + 5)}}.\frac{{x - 5}}{x}\\
 = \frac{{x + 5}}{x} = 1 + \frac{5}{x}\\
b.M = \frac{1}{{20}}x + 1\\
 \leftrightarrow 1 + \frac{5}{x} = \frac{1}{{20}}x + 1\\
 \leftrightarrow 20.5 = x.x\\
 \leftrightarrow 100 = {x^2}\\
 \leftrightarrow x =  \pm 10(tm)\\
\end{array}\)

c. Để M đạt giá trị nguyên <-> 5 chia hết cho x

-> x∈{-5,-1,1,5}