148° Вправа 148Алгебра 8 клас Тарасенкова 2025 · § 4 · стор. 47

148. Спростіть вираз.

1) ${\displaystyle \frac{1}{1+a}+\frac{a}{1+a}=}$${\displaystyle \frac{1+a}{1+a}=}$${\displaystyle 1}$;

2) ${\displaystyle \frac{4}{2-y}-\frac{2y}{2-y}=}$${\displaystyle \frac{4-2y}{2-y}=}$${\displaystyle \frac{2\cancel{\left(2-y\right)}}{\cancel{2-y}}=}$${\displaystyle 2}$;

3) ${\displaystyle \frac{3b}{b-1}+\frac{1-4b}{b-1}=}$${\displaystyle \frac{3b+1-4b}{b-1}=}$${\displaystyle \frac{1-b}{b-1}=}$${\displaystyle -1}$;

4) ${\displaystyle \frac{16}{4+y}-\frac{y^2}{y+4}=}$${\displaystyle \frac{16-y^2}{y+4}=}$${\displaystyle \frac{(4-y)\cancel{\left(4+y\right)}}{\cancel{4+y}}=}$${\displaystyle 4-y}$;

5) ${\displaystyle \frac{a}{a^2-64}+\frac{8}{a^2-64}=}$${\displaystyle \frac{a+8}{(a-8)(a+8)}=}$${\displaystyle \frac{\cancel{a+8}}{(a-8)\cancel{\left(a+8\right)}}=}$${\displaystyle \frac{1}{a-8}}$;

6) ${\displaystyle \frac{-5+c^2}{c-2}-\frac{1}{2-c}=}$${\displaystyle \frac{c^2-5}{c-2}+\frac{1}{c-2}=}$${\displaystyle \frac{c^2-4}{c-2}=}$${\displaystyle \frac{\cancel{\left(c-2\right)}(c+2)}{\cancel{c-2}}=}$${\displaystyle c+2}$;

7) ${\displaystyle \frac{n^2}{3n-33}+\frac{121}{33-3n}=}$${\displaystyle \frac{n^2}{3n-33}-\frac{121}{3n-33}=}$${\displaystyle \frac{n^2-121}{3(n-11)}=}$${\displaystyle \frac{\cancel{\left(n-11\right)}(n+11)}{3\cancel{\left(n-11\right)}}=}$${\displaystyle \frac{n+11}{3}}$.

8) ${\displaystyle 4y-y^2-4=}$${\displaystyle -(y^2-4y+4)=}$${\displaystyle -{(y-2)}^2}$; ${\displaystyle y^2-4y+4=}$${\displaystyle {(y-2)}^2}$;

${\displaystyle \frac{y^2-4y}{y^2-4y+4}-\frac{2y}{4y-y^2-4}=}$${\displaystyle \frac{y^2-4y}{{(y-2)}^2}+\frac{2y}{{(y-2)}^2}}$;${\displaystyle \frac{y^2-4y+2y}{{(y-2)}^2}=}$${\displaystyle \frac{y^2-2y}{{(y-2)}^2}=}$${\displaystyle \frac{y\cancel{\left(y-2\right)}}{\cancel{\left(y-2\right)}(y-2)}=}$${\displaystyle \frac{y}{y-2}}$.