69 Вправа 69Алгебра 8 клас Тарасенкова 2025 · § 1 · стор. 21

69. Визначте ОДЗ кожної змінної виразу.

Розкладаємо дільник на:

1) ${\displaystyle \frac{2a+1}{4-a^2}}$;

${\displaystyle 4-a^2=}$${\displaystyle 0}$;

${\displaystyle (2-a)(2+a)=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 2}$ або ${\displaystyle a=}$${\displaystyle -2}$;

ОДЗ: ${\displaystyle a\ne -2}$, ${\displaystyle a\ne 2}$;

2) ${\displaystyle \frac{3b-2}{9-b^2}}$;

${\displaystyle 9-b^2=}$${\displaystyle 0}$;

${\displaystyle (3-b)(3+b)=}$${\displaystyle 0}$;

${\displaystyle b=}$${\displaystyle 3}$ або ${\displaystyle b=}$${\displaystyle -3}$;

ОДЗ: ${\displaystyle b\ne -3}$, ${\displaystyle b\ne 3}$;

3) ${\displaystyle 3xy:(x^2-16)}$;

${\displaystyle x^2-16=}$${\displaystyle 0}$;

${\displaystyle (x-4)(x+4)=}$${\displaystyle 0}$;

${\displaystyle x=}$${\displaystyle 4}$ або ${\displaystyle x=}$${\displaystyle -4}$;

ОДЗ: ${\displaystyle x\ne -4}$, ${\displaystyle x\ne 4}$; ${\displaystyle y}$ — будь-яке число;

4) ${\displaystyle \frac{2a}{a^2-2a+1}}$;

${\displaystyle a^2-2a+1=}$${\displaystyle 0}$;

${\displaystyle {(a-1)}^2=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1}$;

ОДЗ: ${\displaystyle a\ne 1}$;

5) ${\displaystyle \frac{3b}{4(b^2-6b+9)}}$;

${\displaystyle 4(b^2-6b+9)=}$${\displaystyle 0}$;

${\displaystyle {(b-3)}^2=}$${\displaystyle 0}$;

${\displaystyle b=}$${\displaystyle 3}$;

ОДЗ: ${\displaystyle b\ne 3}$;

6) ${\displaystyle (5-x^2):(25+10x^2+x^4)}$;

${\displaystyle 25+10x^2+x^4=}$${\displaystyle 0}$;

${\displaystyle {(5+x^2)}^2=}$${\displaystyle 0}$;

${\displaystyle x^2=}$${\displaystyle -5}$ — коренів немає;

ОДЗ: ${\displaystyle x}$ — будь-яке число.