\(\frac{x}{x+1}+\frac{x+1}{x+2}+\frac{x+2}{x}=\frac{25}{6}\)

<=> 6x²(x + 2) + 6x(x + 1)² + 6(x + 2)²(x + 1) = 25x(x + 1)(x + 2)

<=> 18x² + 54x² + 54x + 24 = 25x³ + 75x² + 50x

<=> 18x² + 54x² + 54x + 24 - 25x² - 75x² - 50x = 0

<=> -7x³ - 21x² + 4x + 24 = 0

<=> (-7x² - 28x - 24)(x - 1) = 0

vì 7x² + 28x + 24 khác 0 nên:

<=> x - 1 = 0

<=> x = 0

a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm5\end{cases}}\)

\(M=\left(\frac{x}{x+5}-\frac{5}{5-x}+\frac{10x}{x^2-25}\right)\cdot\left(1-\frac{5}{x}\right)\)

\(\Leftrightarrow M=\frac{x^2-5x+5x+25+10x}{\left(x+5\right)\left(x-5\right)}\cdot\frac{x-5}{x}\)

\(\Leftrightarrow M=\frac{\left(x^2+10x+25\right)\left(x-5\right)}{\left(x+5\right)\left(x-5\right)x}\)

\(\Leftrightarrow M=\frac{\left(x+5\right)^2}{x\left(x+5\right)}\)

\(\Leftrightarrow M=\frac{x+5}{x}\)

b) Để \(M\inℤ\)

\(\Leftrightarrow x+5⋮x\)

\(\Leftrightarrow5⋮x\)

\(\Leftrightarrow x\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Mà \(x\ne\pm5\)

\(\Leftrightarrow x\in\left\{1;-1\right\}\)

Vậy để \(M\inℤ\Leftrightarrow x\in\left\{1;-1\right\}\)

\(M=\left(\frac{x}{x+5}-\frac{5}{5-x}+\frac{10x}{x^2-25}\right)\cdot\left(1-\frac{5}{x}\right)\left(x\ne\pm5;x\ne0\right)\)

\(\Leftrightarrow M=\left(\frac{x}{x+5}+\frac{5}{x-5}+\frac{10x}{\left(x-5\right)\left(x+5\right)}\right)\cdot\frac{x-5}{x}\)

\(\Leftrightarrow M=\left(\frac{x^2-5x}{\left(x-5\right)\left(x+5\right)}+\frac{5x+25}{\left(x-5\right)\left(x+5\right)}+\frac{10x}{\left(x-5\right)\left(x+5\right)}\right)\cdot\frac{x-5}{x}\)

\(\Leftrightarrow M=\frac{x^2-5x+5x+25+10x}{\left(x-5\right)\left(x+5\right)}\cdot\frac{x-5}{x}\)

\(\Leftrightarrow M=\frac{x^2+10x+25}{\left(x-5\right)\left(x+5\right)}\cdot\frac{x-5}{x}\)

\(\Leftrightarrow M=\frac{\left(x+5\right)^2\left(x-5\right)}{\left(x-5\right)\left(x+5\right)x}=\frac{x+5}{x}\)

b) M là số nguyên thì x+5 chia hết cho x

=> 5 chia hết cho x

x nguyên => x thuộc Ư (5)={-5;-1;1;5}
Vậy x={-5;-1;1;5} thì M là số nguyên

yêu cầu đầu bài là gì ạ ?