Question

помогите решить даю 20 баллов

Answer 1

Решение во вложении:

Answer 2

$(\frac{r^2-5r+25}{25r^2-1}*\frac{5r^2+r}{r^3+125}-\frac{r+5}{5r^2-r}):\frac{7}{r^2+5r}-\frac{25r+22}{7-35r}=$

$=(\frac{r^2-5r+25}{(5r-1)(5r+1)}*\frac{r(5r+1)}{(r+5)(r^2-5r+25)}-\frac{r+5}{r(5r-1)}):\frac{7}{r^2+5r}-\frac{25r+22}{7-35r}=$

$=(\frac{(r^2-5r+25)*r*(5r+1)}{(5r-1)(5r+1)(r+5)(r^2-5r+25)}-\frac{r+5}{r(5r-1)}):\frac{7}{r^2+5r}-\frac{25r+22}{7-35r}=$

$=(\frac{r}{(5r-1)(r+5)}-\frac{r+5}{r(5r-1)}):\frac{7}{r^2+5r}-\frac{25r+22}{7-35r}=$

$=\frac{r^2-(r+5)(r+5)}{r(5r-1)(r+5)}:\frac{7}{r^2+5r}-\frac{25r+22}{7-35r}=$

$=\frac{r^2-r^2-10r-25}{r(5r-1)(r+5)}:\frac{7}{r^2+5r}-\frac{25r+22}{7-35r}=$

$=\frac{-10r-25}{r(5r-1)(r+5)}:\frac{7}{r^2+5r}-\frac{25r+22}{7-35r}=$

$=\frac{10r+25}{r(1-5r)(r+5)}*\frac{r^2+5r}{7}-\frac{25r+22}{7-35r}=$

$=\frac{(10r+25)*r*(r+5)}{r*(1-5r)(r+5)*7}-\frac{25r+22}{7-35r}=$

$=\frac{10r+25}{(1-5r)*7}-\frac{25r+22}{7-35r}=$

$=\frac{10r+25}{7-35r}-\frac{25r+22}{7-35r}=$

$=\frac{10r+25-25r-22}{7-35r}=\frac{3-15r}{7-35r}=\frac{3*(1-5r)}{7*(1-5r)}=\frac{3}{7}$

Ответ:  $\frac{3}{7}$