87 Вправа 87Алгебра 8 клас Бевз 2025 · § 2 · стор. 22

87. Скороти дріб і знайди його значення.

а) ${\displaystyle \frac{4a^2+12ab+9b^2-1}{2a+3b+1}}$;

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

якщо ${\displaystyle a=}$${\displaystyle \frac{1}{2}}$;

${\displaystyle b=}$${\displaystyle -\frac{2}{3}}$, то ${\displaystyle 2\cdot \frac{1}{2}+3\cdot (-\frac{2}{3})-1}$=

=${\displaystyle 1-2-1=}$${\displaystyle -2}$;

б) ${\displaystyle \frac{(a^4+b^4)(a^3-a^{2}b+a{b}^2-b^3)}{a^8-b^8}}$;

${\displaystyle a^3-a^{2}b+a{b}^2-b^3=}$${\displaystyle a^2(a-b)+b^2(a-b)}$=

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

${\displaystyle a^8-b^8=}$${\displaystyle (a-b)(a+b)(a^2+b^2)(a^4+b^4)}$;

${\displaystyle \frac{\cancel{\left(a^4+b^4\right)}\cancel{\left(a-b\right)}\cancel{\left(a^2+b^2\right)}}{\cancel{\left(a-b\right)}(a+b)\cancel{\left(a^2+b^2\right)}\cancel{\left(a^4+b^4\right)}}=}$${\displaystyle \frac{1}{a+b}}$;

якщо ${\displaystyle a=}$${\displaystyle 12,25}$;

${\displaystyle b=}$${\displaystyle -2\frac{1}{4}}$, то ${\displaystyle \frac{1}{12,25-2,25}}$=

=${\displaystyle \frac{1}{10}=}$${\displaystyle 0,1}$.