Question

((5с^2-c/25c^2-10c+1)+(4/1-25c^2)):(1-(3/5c-1))-(c/5c+1)
надо сократить срочно

Answer 1

$displaystylemathtt{frac{frac{5c^2-c}{25c^2-10c+1}+frac{4}{1-25c^2}}{1-frac{3}{5c-1}}-frac{c}{5c+1}=frac{frac{c(5c-1)}{(5c-1)^2}-frac{4}{25c^2-1}}{frac{25c^2-1-3(5c+1)}{25c^2-1}}-frac{c}{5c+1}=}\\displaystylemathtt{frac{frac{c(5c+1)-4}{25c^2-1}}{frac{25c^2-1-15c-3}{25c^2-1}}-frac{c}{5c+1}=frac{frac{5c^2+c-4}{25c^2-1}}{frac{25c^2-15c-4}{25c^2-1}}-frac{c}{5c+1};}$

сократив знаменатели, мы получаем: 

$\\displaystylemathtt{frac{5c^2+c-4}{25c^2-15c-4}-frac{c}{5c+1}=frac{(5c-4)(c+1)}{(5c-4)(5c+1)}-frac{c}{5c+1}=}\\displaystylemathtt{frac{c+1}{5c+1}-frac{c}{5c+1}=frac{c+1-c}{5c+1}=frac{1}{5c+1}=(5c+1)^{-1}}$

Answer 2

1)с(5с-1)/(5с-1)²²4/[(5с-1)(5с+1)]=c/(5c-1)-4/[(5c-1)(5c+1)]=
=(5c²²+c-4)/[(5c-1)(5c+1)=(5c-4)(c+1)/[(5c-1)(5c+1)]
2)1-3/(5c-1)=(5c-1-3)/(5c-1)=(5c-4)/(5c-1)
3)(5c-4)(c+1)/[(5c-1)(5c+1):(5c-4)/(5c-1)=
=(5c-4)(c+1)/[(5c-1)(5c+1)]*(5c-1)/(5c-4)=(c+1)/(5c+1)
4)(c+1)/(5c+1)-c/(5c+1)=(c+1-c)/(5c+1)=1/(5c+1)