Помогите пожалуйста!!! Фото ниже

Question 1

Помогите пожалуйста!!!
Фото ниже

Question 2

В скобках:
1)
$\frac{t^2-5t+25}{(5t-1)(5t+1)} * \frac{t(5t+1) }{(t+5)(t^2-5t+25)} - \frac{t+5}{t(5t-1)}$

2)
$\frac{1}{5t-1} * \frac{1}{t+5} - \frac{t+5}{t(5t-1)}= \frac{t}{(5t-1)(t+5)}- \frac{t+5}{t(5t-1)}$

3) Первую на "Т", вторую на "(Т+5)
$\frac{t*t}{t(5t-1)(t+5)}- \frac{(t+5)(t+5)}{(t+5)t*(5t-1)} = \frac{t^2}{t(t+5)(5t-1)} - \frac{(t+5)^2}{t(t+5)*(5t-1)} = \frac{t^2-(t+5)^2}{t(t+5)(5t-1)}$

4)
$\frac{t^2-(t+5)^2}{t(t+5)(5t-1)} = \frac{t^2-(t^2+10t+25)}{t(t+5)(5t-1)} = \frac{t^2-t^2-10t-25}{t(t+5)(5t-1)} = -\frac{10t+25}{t(t+5)(5t-1)}$

5)
$-\frac{10t+25}{t(t+5)(5t-1)}= - \frac{10t+25}{(t^2+5t)(5t-1)} = -\frac{10t+25}{5t^3-t^2+25t^2-5t} = - \frac{10t+25}{5t^3+24t^2-5t}$

6)
- \frac{10t+25}{5t^3+24t^2-5t}
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Деление при открытии, заменяем умножением:
1)
$- \frac{10t+25}{5t^3+24t^2-5t} * \frac{t^2+5t}{4} = -\frac{10t+25}{t(5t^2-24t-5)}* \frac{t(t+5)}{4} = - \frac{10t+25}{5t(t+5)-(t+5)}* \frac{t+5}{4}$

2)
$- \frac{10t+25}{5t(t+5)-(t+5)}* \frac{t+5}{4} = -\frac{10t+25}{(5t-1)(t+5)} * \frac{t+5}{4} =- \frac{10t+25}{5t-1} * \frac{1}{4} = -\frac{10t+25}{4(5t-1)}$

3)
$-\frac{10t+25}{20t-4}$
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Вычитание:
1)
$-\frac{10t+25}{20t-4}- \frac{25t+22}{4-20t} =-\frac{10t+25}{20t-4}- \frac{25t+22}{-(20t-4)} = -\frac{10t+25}{20t-4}+\frac{25t+22}{(20t-4)}$

2)
$-\frac{10t+25}{20t-4}+\frac{25t+22}{(20t-4)}= \frac{-(10t+25)+25t+22}{20t-4} = \frac{-10t-25+25t+22}{4(5t-1)} = \frac{15t-3}{4(5t-1)}$

3)
$\frac{15t-3}{4(5t-1)} = \frac{3(5t-1)}{4(5t-1)} = \frac{3}{4}$

Ответ: 3/4 = 0.75