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

1)с(5с-1)/(5с-1)4/[(5с-1)(5с+1)]=c/(5c-1)-4/[(5c-1)(5c+1)]=
=(5c+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)

Тимур Совостенко · Answer 2 · 2019-03-21 10:54:12+3:00

$\displaystyle\mathtt\frac\frac5c^2-c25c^2-10c+1+\frac41-25c^21-\frac35c-1-\fracc5c+1=\frac\fracc(5c-1)(5c-1)^2-\frac425c^2-1\frac25c^2-1-3(5c+1)25c^2-1-\fracc5c+1=\\\\\displaystyle\mathtt\frac\fracc(5c+1)-425c^2-1\frac25c^2-1-15c-325c^2-1-\fracc5c+1=\frac\frac5c^2+c-425c^2-1\frac25c^2-15c-425c^2-1-\fracc5c+1;$

сократив знаменатели, мы получаем:

$\\\\\displaystyle\mathtt\frac5c^2+c-425c^2-15c-4-\fracc5c+1=\frac(5c-4)(c+1)(5c-4)(5c+1)-\fracc5c+1=\\\\\displaystyle\mathtt\fracc+15c+1-\fracc5c+1=\fracc+1-c5c+1=\frac15c+1=(5c+1)^-1$

О проекте

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

Добро пожаловать!