Thực hiện các phép tính sau:

a) $\frac{5x-10}{x^2+7}:(2x-4);$                                                   b) $(x^2-25):\frac{2x+10}{3x-7}$;

c) $\frac{x^2+x}{5x^2-10x+5}:\frac{3x+3}{5x-5}.$

Giải

a) $\frac{5x-10}{x^2+7}:(2x-4)=\frac{5x-10}{x^2+7}.\frac{1}{2x-4}=\frac{5x-10}{\left(x^2+7\right)2\left(x-2\right)}=\frac{5(x-2)}{\left(x^2+7\right)2\left(x-2\right)}=\frac{5}{2\left(x^2+7\right)}$

b) $(x^2-25):\frac{2x+10}{3x-7}=\frac{x^2-25}{1}.\frac{3x-7}{2x+10}=\frac{\left(x-5\right)\left(x+5\right).(3x-7)}{2(x+5)}=\frac{\left(x-5\right)(3x-7)}{2}=\frac{3x^2-22x+35}{2}$

c) $\frac{x^2+x}{5x^2-10x+5}:\frac{3x+3}{5x-5}=\frac{x^2+x}{5x^2-10x+5}.\frac{5x-5}{3x+3}=\frac{x(x+1).5(x-5)}{5{(x-1)}^2.3(x+1)}=\frac{x}{3(x-1)}$