1 Вправа 1 (повторення)Алгебра 8 клас Тарасенкова 2025 · Повторення · стор. 316

1. Визначте ОДЗ кожної змінної виразу:

1) ${\displaystyle \frac{2x-1}{2x^2+1}}$

${\displaystyle 2x^2+1\ne 0}$; оскільки ${\displaystyle 2x^2\ge 0}$, то ${\displaystyle 2x^2+1\ge 1}$;

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

2) ${\displaystyle \frac{3y-2}{4y^2+12y}}$

${\displaystyle 4y^2+12y\ne 0}$;

${\displaystyle 4y(y+3)\ne 0}$;

${\displaystyle 4y\ne 0}$ або ${\displaystyle y+3\ne 0}$;

ОДЗ: ${\displaystyle y\ne 0}$; ${\displaystyle y\ne -3}$;

3) ${\displaystyle \frac{5a+3}{a^2+3a+2}}$

${\displaystyle a^2+3a+2\ne 0}$;

${\displaystyle D=}$${\displaystyle 3^2-4\cdot 1\cdot 2=}$${\displaystyle 9-8=}$${\displaystyle 1}$;

${\displaystyle a_1=}$${\displaystyle \frac{-3-1}{2}=}$${\displaystyle -2}$;

${\displaystyle a_2=}$${\displaystyle \frac{-3+1}{2}=}$${\displaystyle -1}$;

${\displaystyle (a+2)(a+1)\ne 0}$;

ОДЗ: ${\displaystyle a\ne -2}$; ${\displaystyle a\ne -1}$;

4) ${\displaystyle (3x-2y):4(x+1){(2y-1)}^2}$

${\displaystyle 4(x+1){(2y-1)}^2\ne 0}$;

${\displaystyle x+1\ne 0}$ і ${\displaystyle {(2y-1)}^2\ne 0}$;

${\displaystyle x\ne -1}$ і ${\displaystyle 2y\ne 1}$;

ОДЗ: ${\displaystyle x\ne -1}$; ${\displaystyle y\ne 0,5}$;

5) ${\displaystyle \frac{3abc}{{(3a+1)}^2{(4b-1)}^2{c}^3}}$

${\displaystyle {(3a+1)}^2\ne 0}$; ${\displaystyle {(4b-1)}^2\ne 0}$; ${\displaystyle c^3\ne 0}$;

${\displaystyle 3a\ne -1}$; ${\displaystyle 4b\ne 1}$; ${\displaystyle c\ne 0}$;

ОДЗ: ${\displaystyle a\ne -\frac{1}{3}}$; ${\displaystyle b\ne 0,25}$; ${\displaystyle c\ne 0}$;

6) ${\displaystyle \frac{5(x+1)y{z}^2}{2x^3(3-2y){(5z-13)}^2}}$

${\displaystyle x^3\ne 0}$; ${\displaystyle 3-2y\ne 0}$; ${\displaystyle {(5z-13)}^2\ne 0}$;

${\displaystyle x\ne 0}$; ${\displaystyle 2y\ne 3}$; ${\displaystyle 5z\ne 13}$;

ОДЗ: ${\displaystyle x\ne 0}$; ${\displaystyle y\ne 1,5}$; ${\displaystyle z\ne 2,6}$.