68 Вправа 68Алгебра 8 клас Тарасенкова 2025 · § 1 · стор. 21

68. Визначте ОДЗ кожної змінної виразу.

Прирівнюємо дільник до нуля. Добуток дорівнює нулю, якщо хоча б один із:

1) ${\displaystyle \frac{2a+c}{2ac}}$;

${\displaystyle 2ac=}$${\displaystyle 0}$; ${\displaystyle a=}$${\displaystyle 0}$ або ${\displaystyle c=}$${\displaystyle 0}$;

ОДЗ: ${\displaystyle a\ne 0}$, ${\displaystyle c\ne 0}$;

2) ${\displaystyle \frac{3b-a}{4a{b}^2}}$;

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

ОДЗ: ${\displaystyle a\ne 0}$, ${\displaystyle b\ne 0}$;

3) ${\displaystyle (3x+y):4x{y}^3}$;

${\displaystyle 4x{y}^3=}$${\displaystyle 0}$; ${\displaystyle x=}$${\displaystyle 0}$ або ${\displaystyle y=}$${\displaystyle 0}$;

ОДЗ: ${\displaystyle x\ne 0}$, ${\displaystyle y\ne 0}$;

4) ${\displaystyle \frac{2ac}{2(a-1)(c+1)}}$;

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

${\displaystyle a-1=}$${\displaystyle 0}$ або ${\displaystyle c+1=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1}$ або ${\displaystyle c=}$${\displaystyle -1}$;

ОДЗ: ${\displaystyle a\ne 1}$, ${\displaystyle c\ne -1}$;

5) ${\displaystyle 2x^{2}y:\left((3-x)(y-4)\right)}$;

${\displaystyle (3-x)(y-4)=}$${\displaystyle 0}$;

${\displaystyle 3-x=}$${\displaystyle 0}$ або ${\displaystyle y-4=}$${\displaystyle 0}$;

${\displaystyle x=}$${\displaystyle 3}$ або ${\displaystyle y=}$${\displaystyle 4}$;

ОДЗ: ${\displaystyle x\ne 3}$, ${\displaystyle y\ne 4}$;

6) ${\displaystyle \frac{3ac}{{(a+1)}^2{c}^3}}$;

${\displaystyle {(a+1)}^2{c}^3=}$${\displaystyle 0}$;

${\displaystyle a+1=}$${\displaystyle 0}$ або ${\displaystyle c=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle -1}$ або ${\displaystyle c=}$${\displaystyle 0}$;

ОДЗ: ${\displaystyle a\ne -1}$, ${\displaystyle c\ne 0}$;

7) ${\displaystyle \frac{5bc}{2{(b-5)}^2(1-c)}}$;

${\displaystyle 2{(b-5)}^2(1-c)=}$${\displaystyle 0}$;

${\displaystyle b-5=}$${\displaystyle 0}$ або ${\displaystyle 1-c=}$${\displaystyle 0}$;

${\displaystyle b=}$${\displaystyle 5}$ або ${\displaystyle c=}$${\displaystyle 1}$;

ОДЗ: ${\displaystyle b\ne 5}$, ${\displaystyle c\ne 1}$;

8) ${\displaystyle \frac{7ac}{3{(2-a)}^2(c+2)}}$;

${\displaystyle 3{(2-a)}^2(c+2)=}$${\displaystyle 0}$;

${\displaystyle 2-a=}$${\displaystyle 0}$ або ${\displaystyle c+2=}$${\displaystyle 0}$;

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

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

9) ${\displaystyle \frac{5c}{6(4b-1)(1-4c)}}$;

${\displaystyle 6(4b-1)(1-4c)=}$${\displaystyle 0}$;

${\displaystyle 4b-1=}$${\displaystyle 0}$ або ${\displaystyle 1-4c=}$${\displaystyle 0}$;

${\displaystyle 4b=}$${\displaystyle 1}$ або ${\displaystyle 4c=}$${\displaystyle 1}$;

${\displaystyle b=}$${\displaystyle 0,25}$ або ${\displaystyle c=}$${\displaystyle 0,25}$;

ОДЗ: ${\displaystyle b\ne 0,25}$, ${\displaystyle c\ne 0,25}$;

10) ${\displaystyle 2cd:\left((3d-4)(5-3c)\right)}$;

${\displaystyle (3d-4)(5-3c)=}$${\displaystyle 0}$;

${\displaystyle 3d-4=}$${\displaystyle 0}$ або ${\displaystyle 5-3c=}$${\displaystyle 0}$;

${\displaystyle 3d=}$${\displaystyle 4}$ або ${\displaystyle 3c=}$${\displaystyle 5}$;

${\displaystyle d=}$${\displaystyle \frac{4}{3}=}$${\displaystyle 1\frac{1}{3}}$ або ${\displaystyle c=}$${\displaystyle \frac{5}{3}=}$${\displaystyle 1\frac{2}{3}}$;

ОДЗ: ${\displaystyle d\ne 1\frac{1}{3}}$, ${\displaystyle c\ne 1\frac{2}{3}}$.