84 Вправа 84Алгебра 8 клас Бевз 2025 · § 2 · стор. 22

84. Prove the identity.

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

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

${\displaystyle \frac{1}{2}=}$${\displaystyle \frac{1}{2}}$ — тотожність доведено;

b) ${\displaystyle \frac{a^2+ab}{a+b}=}$${\displaystyle \frac{a^3-a}{(a-1)(a+1)}}$;

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

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

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

c) ${\displaystyle \frac{x^3-1}{{(x+1)}^2-x}=}$${\displaystyle \frac{x-2x^2+x^3}{x^2-x}}$;

${\displaystyle \frac{x^3-1}{x^2+2x+1-x}=}$${\displaystyle \frac{(x-1)\cancel{\left(x^2+x+1\right)}}{\cancel{x^2+x+1}}=}$${\displaystyle x-1}$;

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

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

d) ${\displaystyle \frac{a^6-2a^3+1}{{(a^2+a+1)}^2}=}$${\displaystyle {(a-1)}^2}$;

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

${\displaystyle {(a-1)}^2=}$${\displaystyle {(a-1)}^2}$ — тотожність доведено.