Завдання № 3.10

№ 3.10 Алгебра =  № 3.10 Математика

Виконайте дію:

1)$\ \frac{x^2}{x\ -\ 5}\ -\ \frac{25}{x\ -\ 5};$    

2)$\ \frac{36}{y\ +\ 6}\ -\ \frac{y^2}{y\ +\ 6};$ 

3)$\ \frac{x\ -\ 3}{x^2\ -\ 9}\ +\ \frac{6}{x^2\ -\ 9}; $  

4)$\ \frac{7a\ -\ 1}{a^2\ -\ b^2}\ -\ \frac{7b\ -\ 1}{a^2\ -\ b^2};$      

5)$\ \frac{2x\ +\ y}{\left(x\ -\ y\right)^2}\ +\ \frac{x\ -\ 4y}{\left(x\ -\ y\right)^2};$   

6)$\ \frac{9m\ +\ 5n}{\left(m+n\right)^2}\ -\ \frac{m\ -\ 3n}{\left(m+n\right)^2}.$

Розв'язок:

1)$\ \frac{x^2}{x\ -\ 5}\ -\ \frac{25}{x\ -\ 5}\ =\ \frac{x^2\ -\ 25}{x\ -\ 5}\ =$

$= \ \frac{(x\ -\ 5)(x\ +\ 5)}{x\ -\ 5}\ =\ x\ +\ 5;$   

2)$\ \frac{36}{y\ +\ 6}\ -\ \frac{y^2}{y\ +\ 6}\ =\ \frac{36\ -\ y^2}{y\ +\ 6}\ =$

$= \ \frac{(6\ -\ y)(6\ +\ y)}{6\ +\ y}\ =\ 6\ – y; $  

3)$\ \frac{x - 3}{x^2 - 9}\ +\ \frac{6}{x^2 - 9}\ =$

$= \ \frac{x - 3 + 6 }{x^2 - 9}\ =$

$= \ \frac{x\ +\ 3}{x^2\ -\ 9}\ =\ \frac{x\ +\ 3}{(x\ -\ 3)(x\ +\ 3)}\ =\ \frac{1}{x\ -\ 3};$   

4)$\ \frac{7a\ -\ 1}{a^2\ -\ b^2}\ -\ \frac{7b\ -\ 1}{a^2\ -\ b^2}\ =$

$= \ \frac{7a\ -\ 1\ -\ (7b\ -\ 1)}{a^2\ -\ b^2}\ =$

$= \ \frac{7a\ -\ 1\ -\ 7b\ +\ 1}{a^2\ -\ b^2}\ =$

$= \ \frac{7a\ -\ 7b}{a^2\ -\ b^2}\ =\ \frac{7(a\ -\ b)}{(a\ -\ b)(a\ +\ b)}\ =$

$= \ \frac{7}{a\ +\ b};$ 

5)$\ \frac{2x\ +\ y}{\left(x\ -\ y\right)^2}\ +\ \frac{x\ -\ 4y}{\left(x\ -\ y\right)^2}\ =$

$= \ \frac{2x\ +\ y\ +\ x\ -\ 4y}{\left(x\ -\ y\right)^2}\ =$

$ = \ \frac{3x\ -\ 3y}{\left(x\ -\ y\right)^2}\ =\ \frac{3(x\ -\ y)}{\left(x\ -\ y\right)^2}\ =$

$= \ \frac{3}{x\ -\ y};$    

6)$ \frac{9m\ +\ 5n}{\left(m+n\right)^2}\ -\ \frac{m\ -\ 3n}{\left(m+n\right)^2}\ =$

$= \ \frac{9m\ +\ 5n\ -\ (m\ -\ 3n)}{\left(m+n\right)^2}\ =$

$= \ \frac{8m\ +\ 8n}{\left(m+n\right)^2}\ =$

$= \ \frac{8(m\ +\ n)}{\left(m+n\right)^2}\ =\ \frac{8}{m\ +\ n}.$