83 Вправа 83Алгебра 8 клас Бевз 2025 · § 2 · стор. 22

83. Доведи тотожність.

а) ${\displaystyle \frac{4ab}{{(a+b)}^2-{(a-b)}^2}=}$${\displaystyle 1}$;

${\displaystyle \frac{4ab}{a^2+2ab+b^2-(a^2-2ab+b^2)}=}$${\displaystyle \frac{4ab}{a^2+2ab+b^2-a^2+2ab-b^2}}$=

=${\displaystyle \frac{4ab}{4ab}=}$${\displaystyle 1}$;

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

б) ${\displaystyle \frac{x+y}{x^2-y^2}=}$${\displaystyle \frac{x-y}{x^2-2xy+y^2}}$;

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

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

${\displaystyle \frac{1}{x-y}=}$${\displaystyle \frac{1}{x-y}}$ — тотожність доведено;

в) ${\displaystyle \frac{3a^2+2a}{6+9a}=}$${\displaystyle \frac{2(a^3+a)}{6+6a^2}}$;

${\displaystyle \frac{3a^2+2a}{6+9a}=}$${\displaystyle \frac{a\cancel{\left(3a+2\right)}}{3\cancel{\left(2+3a\right)}}=}$${\displaystyle \frac{a}{3}}$;

${\displaystyle \frac{2(a^3+a)}{6+6a^2}=}$${\displaystyle \frac{\cancel{2}a\cancel{\left(a^2+1\right)}}{\cancel{2}\cdot 3\cancel{\left(1+a^2\right)}}=}$${\displaystyle \frac{a}{3}}$;

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

г) ${\displaystyle \frac{a^4-2a^2+1}{a^3-a^2-a+1}=}$${\displaystyle a+1}$;

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

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