Đáp án+giải thích các bước giải:
a)
ĐKXĐ: $x \neq -5$
$A=\dfrac{3x+15}{x^2+10x+25}$
$=\dfrac{3(x+5)}{(x+5)^2}$
$=\dfrac{3}{x+5}$
$\text{Thay x=10 vào A ta được}$
$\dfrac{3}{15}=\dfrac{1}{5}$
b)
ĐKXĐ: $x \neq ± 3$
$B=\dfrac{x}{x+3}-\dfrac{2x}{3-x}-\dfrac{3x^2+9}{x^2-9}$
$=\dfrac{x(x-3)}{(x-3)(x+3)}+\dfrac{2x(x+3)}{(x-3)(x+3)}-\dfrac{3x^2+9}{(x-3)(x+3)}$
$=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{(x-3)(x+3)}$
$=\dfrac{3x-9}{(x-3)(x+3)}$
$=\dfrac{3(x-3)}{(x-3)(x+3)}$
$=\dfrac{3}{x+3}$
c)
ĐKXĐ: $x=-5$
$P=\dfrac{A}{B}=\dfrac{\dfrac{3}{x+5}}{\dfrac{3}{x+3}}$
$=\dfrac{3}{x+5}:\dfrac{3}{x+3}$
$=\dfrac{3}{x+5}.\dfrac{x+3}{x+5}$
$=\dfrac{x+3}{x+5}$
$=1-\dfrac{2}{x+5}$
$\text{Để P nguyên thì}$
$\text{2 chia hết cho (x+5)}$
$⇒ (x+5) ∈ Ư(2) =${$±1; ±2$}
\(\left[ \begin{array}{l}x+5=1\\x+5=-1\\x+5=2\\x+5=-2\end{array} \right.\)
\(\left[ \begin{array}{l}x=-4\\x=-6\\x=-3\\x=-7\end{array} \right.\) (TMĐK)
$\text{Vậy x ∈}${$-3; -4; -6; -7$} $\text{ thì P nguyên}$
