Question

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

Answer 1

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

сократив знаменатели, мы получаем: 

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

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)