mn giúp e vs ạk ($\frac{2x+1}{x^{2}-5x}$ -$\frac{2x}{x^{2}+5x}$) .$\frac{x^{3}-25x}{21x-2}$

Đáp án:

$\dfrac{21x +5}{21x -2}$

Giải thích các bước giải:

$\left(\dfrac{2x+1}{x^2 – 5x} -\dfrac{2x}{x^2 + 5x}\right)\cdot \dfrac{x^3 – 25x}{21x -2}\qquad (*)$

$ĐKXĐ:\begin{cases}x^2-5x \ne 0\\x^2 +5x \ne 0\\21x -2\ne 0\end{cases}\longrightarrow \begin{cases}x \ne 0\\x \ne \pm 5\\x \ne \dfrac{2}{21}\end{cases}$

$(*)=\left[\dfrac{(2x+1)(x+5)}{x(x-5)(x+5)}-\dfrac{2x(x-5)}{x(x-5)(x+5)}\right]\cdot\dfrac{x(x-5)(x+5)}{21x -2}$

$= \dfrac{2x^2 + 11x + 5 – (2x^2 – 10x)}{x(x-5)(x+5)}\cdot\dfrac{x(x-5)(x+5)}{21x -2}$

$= \dfrac{21x +5}{21x -2}$