72 Вправа 72Алгебра 8 клас Бевз 2025 · § 2 · стор. 20

72. Доведи тотожність.

а) ${\displaystyle \frac{5x^{2}y}{xy}=}$${\displaystyle \frac{10xy}{2y}}$;

${\displaystyle \frac{5x^{2}y}{xy}=}$${\displaystyle \frac{5x\cdot \cancel{xy}}{\cancel{xy}}=}$${\displaystyle 5x}$;

${\displaystyle \frac{10xy}{2y}=}$${\displaystyle \frac{5x\cdot \cancel{2y}}{\cancel{2y}}=}$${\displaystyle 5x}$;

${\displaystyle 5x=}$${\displaystyle 5x}$ — тотожність доведено;

б) ${\displaystyle \frac{4b(a+b)}{12b}=}$${\displaystyle \frac{5a(a+b)}{15a}}$;

${\displaystyle \frac{4b(a+b)}{12b}=}$${\displaystyle \frac{\cancel{4b}(a+b)}{\cancel{4b}\cdot 3}=}$${\displaystyle \frac{a+b}{3}}$;

${\displaystyle \frac{5a(a+b)}{15a}=}$${\displaystyle \frac{\cancel{5a}(a+b)}{\cancel{5a}\cdot 3}=}$${\displaystyle \frac{a+b}{3}}$;

${\displaystyle \frac{a+b}{3}=}$${\displaystyle \frac{a+b}{3}}$ — тотожність доведено;

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

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

${\displaystyle a+1=}$${\displaystyle a+1}$ — тотожність доведено;

г) ${\displaystyle \frac{4a^2-x^2}{2a-x}=}$${\displaystyle 2a+x}$;

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

${\displaystyle 2a+x=}$${\displaystyle 2a+x}$ — тотожність доведено;

ґ) ${\displaystyle \frac{{(a+c)}^2}{a+c}=}$${\displaystyle a+c}$;

${\displaystyle \frac{{(a+c)}^2}{a+c}=}$${\displaystyle \frac{(a+c)\cancel{\left(a+c\right)}}{\cancel{a+c}}=}$${\displaystyle a+c}$;

${\displaystyle a+c=}$${\displaystyle a+c}$ — тотожність доведено;

д) ${\displaystyle \frac{x^2+2x+1}{x+1}=}$${\displaystyle x+1}$;

${\displaystyle \frac{x^2+2x+1}{x+1}=}$${\displaystyle \frac{(x+1)\cancel{\left(x+1\right)}}{\cancel{x+1}}=}$${\displaystyle x+1}$;

${\displaystyle x+1=}$${\displaystyle x+1}$ — тотожність доведено.