Dùng quy tắc đổi dấu rồi thực hiện các phép tính:

a) $\frac{4x+13}{5x\left(x-7\right)}-\frac{x-48}{5x\left(x-7\right)};$                                         b) $\frac{1}{x-5x^2}-\frac{25x-15}{25x^2-1}.$

Giải

a) $\frac{4x+13}{5x\left(x-7\right)}-\frac{x-48}{5x\left(x-7\right)}=\frac{4x+13}{5x\left(x-7\right)}+\frac{x-48}{-5x\left(7-x\right)}=\frac{4x+13}{5x\left(x-7\right)}+\frac{x-48}{5x\left(x-7\right)}$

$=\frac{4x+13+x-48}{5x\left(x-7\right)}=\frac{5x-35}{5x\left(x-7\right)}=\frac{5\left(x-7\right)}{5x\left(x-7\right)}=\frac{1}{x}$

b) $\frac{1}{x-5x^2}-\frac{25x-15}{25x^2-1}=\frac{1}{x\left(1-5x\right)}+\frac{25x-15}{-\left(25x^2-1\right)}=\frac{1}{x\left(1-5x\right)}+\frac{25x-15}{1-25x^2}$

$$\begin{gathered} =\frac{1}{x\left(1-5x\right)}+\frac{25x-15}{\left(1-5x\right)\left(1+5x\right)}=\frac{\left(1+5x\right)+x\left(25x-15\right)}{x\left(1-5x\right)\left(1+5x\right)}=\frac{1+5x+25x^2-15x}{x\left(1-5x\right)\left(1+5x\right)} \\ =\frac{1-10x+25x^2}{x\left(1-5x\right)\left(1+5x\right)}=\frac{{\left(1-5x\right)}^2}{x\left(1-5x\right)\left(1+5x\right)}=\frac{1-5x}{x\left(1+5x\right)} \end{gathered}$$