a) ĐKXĐ : $x-5x^2 \neq 0 $
$⇔x.(1-5x) \neq 0 $
$⇔ \left\{ \begin{array}{l}x \neq 0 \\1-5x \neq 0\end{array} \right.$
$⇔ \left\{ \begin{array}{l}x \neq 0 \\x \neq \dfrac{1}{5}\end{array} \right.$
b) $ĐKXĐ : 25x^2-1 \neq 0 $
$⇔(5x-1).(5x+1) \neq 0 $
$⇔ \left\{ \begin{array}{l}5x-1 \neq 0 \neq 0 \\5x+1 \neq 0\end{array} \right.$
$⇔ \left\{ \begin{array}{l}x \neq \dfrac{1}{5}\\x \neq -\dfrac{1}{5} \end{array} \right.$
c) Để $B = 0$
$⇔ \dfrac{25x-15}{25x^2-1} = 0$
$⇒25x-15=0$
$⇔25x=15$
$⇔x=\dfrac{15}{25} = \dfrac{3}{5}$ ( Thỏa mãn $ĐKXĐ$ )
Vậy : $x=\dfrac{3}{5}$ thì $B = 0$
d) Ta có :
$A - B = \dfrac{1}{x-5x^2}-\dfrac{25x-15}{25x^2-1} $
$= \dfrac{1}{x.(1-5x)} -\dfrac{25x-15}{(5x-1).(5x+1)} $
$= \dfrac{1}{x.(1-5x)} + \dfrac{25x-15}{(1-5x).(5x+1)} $
$= \dfrac{5x+1+25x-15}{(1-5x).(5x+1)} $
$= \dfrac{30x-14}{(1-5x).(5x+1)} $
Vậy : $A-B = \dfrac{30x-14}{(1-5x).(5x+1)} $
Chúc bạn học tốt !
